What is the simplest way to find the acute-angle bisector?

Rewrite both line equations so the constant terms are positive, then compute a1 a2 + b1 b2. If this is negative, the plus sign in the bisector formula gives the acute bisector. If positive, plus gives the obtuse bisector.

Points and Straight Lines

Q: How do I quickly decide which side of a line a point is on?

Plug the point into the expression ax + by + c. If the value has the same sign as c, the point is on the origin side; opposite sign means the non-origin side; zero means the point is on the line.

Q: Why does the ratio formula have a minus sign in front?

Because the ratio m:n is positive for internal division and negative for external division. The minus sign ensures internal (points on opposite sides of the line, giving opposite signs for L(P1) and L(P2)) yields a positive ratio.

Q: How is the image formula related to the foot-of-perpendicular formula?

The two are identical except for a factor of 2. The image sits at twice the perpendicular distance from the original point on the far side of the line. If M is the foot of perpendicular from P, then the image is P' = 2M - P.

Q: How do I reflect a line in another line?

Find where the two lines meet - that point is fixed by the reflection. Pick another point on the line to be reflected, find its image in the mirror line using the point-reflection formula, and draw the line through the intersection and this image.

Q: What is the image of a point in another point?

The image of P(x1, y1) in the point (h, k) is (2h - x1, 2k - y1) - because (h, k) must be the midpoint of P and its image. This is a 180 degree rotation about (h, k), not a reflection in a line.

MathsStraight LinesFor JEE aspirants

Once we know the equation of a straight line, we can ask questions about points relative to that line - which side of the line does a given point lie on, how far is it from the line, what is its foot of perpendicular, what is its mirror image? This concept covers all of these point-line relationships: position of a point (or two points) relative to a line, perpendicular distance, foot of perpendicular, image of a point, reflection of a line, angle bisectors of two intersecting lines with acute/obtuse discrimination, and the bisector containing a given point.

Key Formulas - Quick Reference

Perpendicular distance from $P (x_{1}, y_{1})$ to $a x + b y + c = 0$ : $d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$ .
Two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ lie on the same side of $a x + b y + c = 0$ iff $(a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c) > 0$ .
Ratio in which $a x + b y + c = 0$ divides segment $\overline{P_{1} P_{2}}$ : $\frac{m}{n} = - \frac{a x _{1} + b y _{1} + c}{a x _{2} + b y _{2} + c}$ .
Foot of perpendicular from $(x_{1}, y_{1})$ to $a x + b y + c = 0$ : $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = - \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}$ .
Image of $(x_{1}, y_{1})$ in $a x + b y + c = 0$ : $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = - 2 \cdot \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}$ .
Image of $(x_{1}, y_{1})$ in the point $(h, k)$ : $(2 h - x_{1}, 2 k - y_{1})$ .
Angle bisectors of $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ : $\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = \pm \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}}$ .

1. Position of a Point Relative to a Line

Consider the line $L : a x + b y + c = 0$ and a point $P (x_{1}, y_{1})$ not on it. The value of $L (P) = a x_{1} + b y_{1} + c$ tells us which side of the line $P$ lies on:

If $L (P)$ has the same sign as $c$ , then $P$ lies on the same side of the line as the origin - the "origin side".
If $L (P)$ has the opposite sign to $c$ , then $P$ lies on the non-origin side of the line.
If $L (P) = 0$ , then $P$ lies on the line.

Two points relative to a line

Two points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ lie

on the same side of $a x + b y + c = 0$ iff $(a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c) > 0$ ;
on opposite sides iff $(a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c) < 0$ .

Intuition. $a x + b y + c$ is a continuous function of $(x, y)$ that vanishes exactly on the line. It has one sign on one side and the opposite sign on the other. Two points on the same side give the same sign; on opposite sides, opposite signs.

Figure 1. The sign of

L (x, y) = a x + b y + c

is constant on each side of the line and zero on it. Two points lie on the same side exactly when

L (P) L (Q) > 0

Solved Example 1

Show that

(1, 4)

and

(0, - 3)

lie on opposite sides of

x + 3 y + 7 = 0

Solution:

At $(1, 4)$ : $L = 1 + 12 + 7 = 20 > 0$ . At $(0, - 3)$ : $L = 0 - 9 + 7 = - 2 < 0$ . Opposite signs $\Rightarrow$ opposite sides.

Solved Example 2

Which of

(1, 1)

(- 1, 2)

(2, 3)

lies on the same side of

4 x + 3 y - 5 = 0

as the origin?

Solution:

At origin: $L = - 5 < 0$ (so "origin side" $=$ negative side).

$(1, 1)$ : $4 + 3 - 5 = 2 > 0$ - opposite side.

$(- 1, 2)$ : $- 4 + 6 - 5 = - 3 < 0$ - same side as origin.

$(2, 3)$ : $8 + 9 - 5 = 12 > 0$ - opposite side.

Answer: $(- 1, 2)$ .

2. Ratio in Which a Line Divides a Segment

Suppose the line $L : a x + b y + c = 0$ meets segment $\overline{P_{1} P_{2}}$ (where $P_{1} = (x_{1}, y_{1})$ , $P_{2} = (x_{2}, y_{2})$ ) at some point $R$ . Let $R$ divide $\overline{P_{1} P_{2}}$ in the ratio $m : n$ . Then

\frac{m}{n} = - \frac{a x _{1} + b y _{1} + c}{a x _{2} + b y _{2} + c} .

Sign convention:

If $P_{1}$ and $P_{2}$ are on opposite sides of the line, the two values $a x_{1} + b y_{1} + c$ and $a x_{2} + b y_{2} + c$ have opposite signs, so $\frac{m}{n} > 0$ - the division is internal.
If $P_{1}$ and $P_{2}$ are on the same side, both values have the same sign, so $\frac{m}{n} < 0$ - the division is external.

Figure 2. The line meets

\overline{A B}

in the ratio

\frac{m}{n} = - \frac{a x _{1} + b y _{1} + c}{a x _{2} + b y _{2} + c}

. A positive ratio means internal division (opposite sides), a negative ratio means external division (same side).

Solved Example 3

Find the ratio in which the line

x + y - 5 = 0

divides the segment joining

A (1, 2)

and

B (- 3, 4)

Solution:

At $A$ : $1 + 2 - 5 = - 2$ . At $B$ : $- 3 + 4 - 5 = - 4$ . Both negative $\Rightarrow$ same side $\Rightarrow$ external division.

$\frac{m}{n} = - \frac{- 2}{- 4} = - \frac{1}{2}$ . So the line divides $\overline{A B}$ in the ratio $1 : 2$ externally.

3. Perpendicular Distance from a Point to a Line

The length of the perpendicular from $P (x_{1}, y_{1})$ to the line $a x + b y + c = 0$ is

d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}} .

The absolute value is essential: distance is non-negative. Without the modulus the expression is the signed distance, positive on one side of the line and negative on the other.

Figure 3. Perpendicular distance.

d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

; without the modulus the same expression is the signed distance, positive on one side and negative on the other.

Distance from origin. Setting

(x_{1}, y_{1}) = (0, 0)

gives

d = \frac{∣ c ∣}{a ^{2} + b ^{2}}

- the perpendicular distance from the origin to

a x + b y + c = 0

Solved Example 4

Find the distance from

(2, 1)

to the line

12 x - 5 y + 9 = 0

Solution:

$d = \frac{∣12 ( 2 ) - 5 ( 1 ) + 9∣}{144 + 25} = \frac{∣24 - 5 + 9∣}{13} = \frac{28}{13}$ .

Solved Example 5

Find all points on

x + y = 4

that lie at unit distance from

4 x + 3 y - 10 = 0

Solution:

A general point on $x + y = 4$ is $(t, 4 - t)$ . Distance to the given line:

$\frac{∣4 t + 3 ( 4 - t ) - 10∣}{16 + 9} = \frac{∣ t + 2∣}{5} = 1 \Rightarrow ∣ t + 2∣ = 5 \Rightarrow t = 3 or t = - 7$ .

Points: $(3, 1)$ and $(- 7, 11)$ .

4. Foot of the Perpendicular

Let $M$ be the foot of the perpendicular from $P (x_{1}, y_{1})$ to the line $a x + b y + c = 0$ . The coordinates of $M$ satisfy

\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = - \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}} .

Why this works. The vector $(a, b)$ is normal to the line, so $P M$ is along $(a, b)$ : hence $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b}$ . The common value comes from requiring $M$ to lie on the line.

Geometric method (equivalent). Write the equation of $P M$ using point-slope (slope $= \frac{b}{a}$ , perpendicular to line's slope $- \frac{a}{b}$ ). Solve $P M$ and the given line simultaneously.

Solved Example 6

Find the foot of the perpendicular from

P (- 3, 5)

x - y + 2 = 0

Solution:

Here $a = 1, b = - 1, c = 2$ ; $x_{1} = - 3, y_{1} = 5$ . Common ratio: $- \frac{- 3 - 5 + 2}{1 + 1} = - \frac{- 6}{2} = 3$ .

$x = x_{1} + a \cdot 3 = - 3 + 3 = 0$ . $y = y_{1} + b \cdot 3 = 5 - 3 = 2$ . Foot: $M (0, 2)$ .

5. Image (Reflection) of a Point in a Line

The image (mirror reflection) of $P (x_{1}, y_{1})$ in the line $a x + b y + c = 0$ has coordinates $(x^{'}, y^{'})$ satisfying

\frac{x ^{'} - x _{1}}{a} = \frac{y ^{'} - y _{1}}{b} = - 2 \cdot \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}} .

The only difference from the foot-of-perpendicular formula is the factor of $2$ - because the image is at twice the perpendicular distance from $P$ , on the other side of the line.

Figure 4. Foot and image share one formula. The foot uses

- \frac{a x _{1} + b y _{1} + c}{a ^{2} + b ^{2}}

while the image uses twice that value, because

M

is the midpoint of

P

and

P^{'}

Geometric method (working rule)

Find the foot of the perpendicular $M$ from $P$ to the line.
$M$ is the midpoint of $P$ and its image $P^{'}$ . Solve for $P^{'}$ : $P^{'} = 2 M - P$ .

Image about a point (as opposed to a line)

The image of $P (x_{1}, y_{1})$ in the point $(h, k)$ is $P^{'} (2 h - x_{1}, 2 k - y_{1})$ - because $(h, k)$ is the midpoint of $P$ and $P^{'}$ .

Figure 5. Reflection in a point. Since

(h, k)

is the midpoint of

P

and

P^{'}

, the image of

P (x_{1}, y_{1})

P^{'} (2 h - x_{1}, 2 k - y_{1})

Solved Example 7

Find the image of

P (- 1, 2)

in the line

2 x - 3 y + 4 = 0

Solution:

Common ratio: $- 2 \cdot \frac{2 ( - 1 ) - 3 ( 2 ) + 4}{4 + 9} = - 2 \cdot \frac{- 4}{13} = \frac{8}{13}$ .

$x^{'} = - 1 + 2 \cdot \frac{8}{13} = - 1 + \frac{16}{13} = \frac{3}{13}$ . $y^{'} = 2 + (- 3) \cdot \frac{8}{13} = 2 - \frac{24}{13} = \frac{2}{13}$ .

Image: $(\frac{3}{13}, \frac{2}{13})$ .

Solved Example 8

Find the image of

(1, 2)

in the

y

-axis.

Solution:

The $y$ -axis is $x = 0$ , i.e., $x + 0 \cdot y + 0 = 0$ . Applying the formula with $a = 1, b = 0, c = 0$ : common ratio $= - 2 \cdot \frac{1}{1} = - 2$ . So $x^{'} = 1 - 2 = - 1$ , $y^{'} = 2$ . Image: $(- 1, 2)$ .

Equivalently, reflection in the $y$ -axis simply flips the sign of $x$ .

Reflection of a line in another line

To reflect the line $L_{1}$ in the mirror line $L_{2}$ , use the fact that reflection preserves the point of intersection (if they meet) and reflects the direction:

Find the point $P$ where $L_{1}$ meets $L_{2}$ (if $L_{1} ∥ L_{2}$ , treat separately - the image is a line parallel to both at equal distance on the other side).
Take any convenient point $Q$ on $L_{1}$ (other than $P$ ) and find its image $Q^{'}$ in $L_{2}$ using the point-reflection formula above.
The image line passes through $P$ and $Q^{'}$ .

Figure 7. Reflecting a line. The point of intersection

P

stays fixed, so reflecting any second point

Q

L_{1}

in the mirror

L_{2}

and joining

P

Q^{'}

gives the image line.

Solved Example 9

Find the image of the line

x + y = 4

in the line

y = x

Solution:

The two lines meet at $(2, 2)$ . Take $Q = (4, 0)$ on the first line. Its image in $y = x$ (reflection in $y = x$ swaps coordinates) is $Q^{'} = (0, 4)$ .

The image line passes through $(2, 2)$ and $(0, 4)$ . Slope $= \frac{4 - 2}{0 - 2} = - 1$ . Equation: $y - 2 = - 1 (x - 2) \Rightarrow x + y = 4$ .

The reflection is the line itself - because $x + y = 4$ is perpendicular to $y = x$ , so it maps to itself.

6. Angle Bisectors of Two Intersecting Lines

Given two intersecting lines $L_{1} : a_{1} x + b_{1} y + c_{1} = 0$ and $L_{2} : a_{2} x + b_{2} y + c_{2} = 0$ , the equations of the two angle bisectors are

\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = \pm \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}} .

Why. Any point on a bisector is equidistant from the two lines. Setting the perpendicular distances equal (with $\pm$ for the two bisectors of the two supplementary angles) gives the formula.

The two bisectors are always perpendicular to each other.

Figure 6. Every point of a bisector is equidistant from both lines, which gives

\frac{a _{1} x + b _{1} y + c _{1}}{a _{1}^{2} + b _{1}^{2}} = \pm \frac{a _{2} x + b _{2} y + c _{2}}{a _{2}^{2} + b _{2}^{2}}

. The two bisectors are always perpendicular.

Discriminating acute vs obtuse angle bisector

Method 1 (via tangent of angle). Take one of the given lines (slope $m_{1}$ ) and one of the two bisectors (slope $m_{b}$ ). Compute $tan ϕ = \frac{m _{1} - m _{b}}{1 + m _{1} m _{b}}$ :

If $∣ tan ϕ ∣ < 1$ , then $ϕ < 4 5^{\circ}$ so the full angle $2 ϕ < 9 0^{\circ}$ : this bisector bisects the acute angle.
If $∣ tan ϕ ∣ > 1$ , then $ϕ > 4 5^{\circ}$ so $2 ϕ > 9 0^{\circ}$ : this bisector bisects the obtuse angle.

Method 2 (sign of $a_{1} a_{2} + b_{1} b_{2}$ ). After writing both line equations with positive constant terms (i.e., $c_{1} > 0$ and $c_{2} > 0$ ; multiply through by $- 1$ if needed):

If $a_{1} a_{2} + b_{1} b_{2} < 0$ : the origin lies in the acute angle between the two lines, so the " $+$ " sign gives the acute bisector.
If $a_{1} a_{2} + b_{1} b_{2} > 0$ : the origin lies in the obtuse angle, so the " $+$ " sign gives the obtuse bisector.

Equivalently, once you know which angle contains the origin (see next subsection), the bisector of that angle is the one that also contains the origin.

Bisector containing a given point

To pick out the bisector that contains a specified point $(α, β)$ :

Compute $u_{1} = a_{1} α + b_{1} β + c_{1}$ and $u_{2} = a_{2} α + b_{2} β + c_{2}$ (the values of the two line expressions at the given point).
If $u_{1}$ and $u_{2}$ have the same sign, then the " $+$ " sign in the bisector formula gives the bisector through $(α, β)$ .
If they have opposite signs, the " $-$ " sign gives it.

Bisector containing the origin is a special case of the above with $(α, β) = (0, 0)$ : it depends on the signs of $c_{1}$ and $c_{2}$ .

Solved Example 10

For the lines

4 x + 3 y - 6 = 0

and

5 x + 12 y + 9 = 0

, find (i) the bisector of the acute angle and (ii) the bisector containing the origin.

Solution:

Both bisectors. $\frac{4 x + 3 y - 6}{5} = \pm \frac{5 x + 12 y + 9}{13}$ .

Taking $+$ : $13 (4 x + 3 y - 6) = 5 (5 x + 12 y + 9) \Rightarrow 52 x + 39 y - 78 = 25 x + 60 y + 45 \Rightarrow 27 x - 21 y - 123 = 0 \Rightarrow 9 x - 7 y - 41 = 0$ .

Taking $-$ : $13 (4 x + 3 y - 6) = - 5 (5 x + 12 y + 9) \Rightarrow 77 x + 99 y - 33 = 0 \Rightarrow 7 x + 9 y - 3 = 0$ .

(i) Method 2 for acute. Rewrite with positive constants: $- 4 x - 3 y + 6 = 0$ and $5 x + 12 y + 9 = 0$ . Now $a_{1} a_{2} + b_{1} b_{2} = (- 4) (5) + (- 3) (12) = - 20 - 36 = - 56 < 0$ . So the origin lies in the acute angle, and the acute bisector is the one containing the origin.

Check origin against the two bisectors: $9 (0) - 7 (0) - 41 = - 41 \neq = 0$ ; $7 (0) + 9 (0) - 3 = - 3 \neq = 0$ - both fail, but we compare the sign of the LHS at origin against $c$ -sign of the bisector line. Simpler: apply the bisector-containing-origin formula. At origin, $4 (0) + 3 (0) - 6 = - 6 < 0$ and $5 (0) + 12 (0) + 9 = 9 > 0$ - opposite signs $\Rightarrow$ use " $-$ " sign. That gives $7 x + 9 y - 3 = 0$ .

So the acute bisector (which contains the origin here) is $7x + 9y - 3 = 0$ , and the obtuse bisector is $9 x - 7 y - 41 = 0$ .

(ii) Bisector containing the origin is $7 x + 9 y - 3 = 0$ (same as above).

Solved Example 11

Find the bisector of the angle between

x + 2 y - 11 = 0

and

3 x - 6 y - 5 = 0

containing the point

(1, - 3)

Solution:

At $(1, - 3)$ : $u_{1} = 1 - 6 - 11 = - 16$ and $u_{2} = 3 + 18 - 5 = 16$ . Opposite signs $\Rightarrow$ use $-$ sign.

Bisector: $\frac{x + 2 y - 11}{1 + 4} = - \frac{3 x - 6 y - 5}{9 + 36}$ , i.e., $\frac{x + 2 y - 11}{5} = - \frac{3 x - 6 y - 5}{3 5}$ .

$3 (x + 2 y - 11) = - (3 x - 6 y - 5) \Rightarrow 3 x + 6 y - 33 + 3 x - 6 y - 5 = 0 \Rightarrow 6 x - 38 = 0 \Rightarrow 3 x - 19 = 0$ .

7. Line Equally Inclined to Two Given Lines

Any line whose slope makes equal angles with two given lines $L_{1}$ and $L_{2}$ must be parallel to one of the two angle bisectors of $L_{1}$ and $L_{2}$ . Hence:

Write the two bisectors of $L_{1}$ and $L_{2}$ using the formula in Section 6.
Any line "equally inclined to $L_{1}$ and $L_{2}$ " and passing through a given point $(x_{0}, y_{0})$ has the slope of one of these two bisectors and passes through $(x_{0}, y_{0})$ .

Figure 8. Equal inclination. A line making equal angles with

L_{1}

and

L_{2}

must be parallel to one of their two bisectors, so take the bisector's slope and push the line through the given point.

Solved Example 12

Find the equation of the line through

(1, 2)

equally inclined to the lines

x - y = 0

and

x + y = 0

Solution:

Bisectors of $x - y = 0$ and $x + y = 0$ : $\frac{x - y}{2} = \pm \frac{x + y}{2}$ , i.e., $x - y = \pm (x + y)$ , giving $y = 0$ and $x = 0$ .

So a line equally inclined to both has slope $0$ or is vertical. Through $(1, 2)$ :

Horizontal: $y = 2$ . Vertical: $x = 1$ . Both are valid answers.

Common Mistakes to Avoid

Watch out

Distance formula sign: always take the absolute value in the numerator - a "negative distance" of $- \frac{28}{13}$ signals you forgot the modulus, not that the point is behind the line.
Same-side / opposite-side test: the criterion is the sign of the product $(a x_{1} + b y_{1} + c) (a x_{2} + b y_{2} + c)$ , not the individual signs. A common error is to compare "both positive" instead of "product positive".
Ratio in which a line divides a segment: the answer is $- \frac{a x _{1} + b y _{1} + c}{a x _{2} + b y _{2} + c}$ . Positive ratio $\Rightarrow$ internal (points on opposite sides). Negative ratio $\Rightarrow$ external (same side).
Image formula factor of 2: the foot of the perpendicular uses the factor $- \frac{L ( P )}{a ^{2} + b ^{2}}$ ; the image uses $- 2 \cdot \frac{L ( P )}{a ^{2} + b ^{2}}$ . Missing the $2$ gives the foot, not the image.
Acute vs obtuse bisector - Method 2: both equations must first be rewritten so their constant terms are positive; only then does the sign of $a_{1} a_{2} + b_{1} b_{2}$ correctly identify which angle contains the origin.
Reflection of a line in a parallel line: the two never meet, so the "find intersection point" step fails. The image is a line parallel to both, at equal distance on the far side. Compute using perpendicular distance directly.

Frequently Asked Questions

Q1. How do I quickly decide which side of a line a point is on?

Plug the point into the expression $a x + b y + c$ . If the value has the same sign as $c$ , the point is on the origin side; opposite sign means the non-origin side; zero means the point is on the line.

Q2. Why does the ratio formula have a minus sign in front?

Because the ratio $m : n$ in which a line divides a segment is positive for internal division (points on opposite sides of the line, so $L (P_{1})$ and $L (P_{2})$ have opposite signs) and negative for external division. The minus sign in $- \frac{L ( P _{1} )}{L ( P _{2} )}$ ensures the sign works out correctly.

Q3. How is the image formula related to the foot-of-perpendicular formula?

The two are identical except for a factor of $2$ : the image sits at twice the perpendicular distance from the original point on the far side of the line. So if the foot of perpendicular from $P$ is $M$ , then the image is $P^{'} = 2 M - P$ .

Q4. Are the two angle bisectors of a pair of lines always perpendicular?

Yes. The two bisectors of the two supplementary angles between any pair of intersecting lines are always perpendicular. This is why picking one bisector automatically fixes the other as its perpendicular.

Q5. What's the simplest way to find the acute-angle bisector?

Rewrite both line equations so the constant terms are positive, then compute $a_{1} a_{2} + b_{1} b_{2}$ . If this is negative, the " $+$ " sign in the bisector formula gives the acute bisector. If positive, the " $+$ " sign gives the obtuse bisector. Simple and mechanical, no trigonometry needed.

Q6. What does "bisector containing the origin" mean?

Of the two angle bisectors of a pair of intersecting lines, one passes through (or its extension covers) the region containing the origin, and the other doesn't. The "bisector containing the origin" is the one whose region includes the origin. It's a way to pin down a specific bisector without invoking acute/obtuse.

Q7. How do I reflect a line in another line?

Find where the two lines meet (the intersection is fixed by any reflection through a line passing through it). Pick another convenient point on the line you want to reflect, find its image in the mirror line using the point-reflection formula, and draw the line through the intersection and this image. That's the reflected line.

Q8. What is the image of a point in another point?

The image of $P (x_{1}, y_{1})$ in the point $(h, k)$ is $(2 h - x_{1}, 2 k - y_{1})$ - because the point $(h, k)$ must be the midpoint of $P$ and its image. This is a $18 0^{\circ}$ rotation about $(h, k)$ , not a reflection in a line.

Previous year questions on Points and Straight Lines

18 questions from past papers, each with a step-by-step solution.

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