Points and Straight Lines

POSITION OF A POINT WITH RESPECT TO A GIVEN LINE

Let the given line be ax + by + c = 0, and we have to check whether the point ($\alpha$, $\beta$) lies above the line or below the line. To check first draw a perpendicular line from P($\alpha$, $\Rightarrow$) on the x-axis which intersect the given line at Q. x-coordinate of Q will be $\alpha$ and corresponding to $\alpha$, y-coordinate of Q will be

${\beta }^{\prime }=-\left(\frac{a\alpha +c}{b}\right)$. Now

if $\beta >{\beta }^{\prime }$, the point will lie above the line

if $\beta ={\beta }^{\prime }$ point will lie on the line

and if$\beta <{\beta }^{\prime }$ point will be lie below the line.

Illustration 1: If the point ($\alpha$, $\alpha$²) lies between the acute angle region formed between the lines y = 2x and y = 3x, then find the range of $\alpha$.

Key concept 1: Since y co-ordinate of ($\alpha$, $\alpha$²) is always non negative, point ($\alpha$, $\alpha$²) will always lie above the x-axis. Now for acute angle region the point ($\alpha$, $\beta$) should lie below the line y = 3x and above the line y = 2x.

Solution: ($\alpha$, $\beta$) should lie below the lie y = 3x $\Rightarrow$ 3$\alpha$ < $\alpha$² and

($\alpha$, $\beta$)should also lie above the line y = 2x $\Rightarrow$ $\alpha$² > 2$\alpha$

$\Rightarrow$ 2$\alpha$ < $\alpha$² < 3$\alpha$ $\Rightarrow$ 2 < $\alpha$ < 3

Key concept 2: Join P with the origin now slope of OP should lie between the slope of the lines y = 2x and y = 3x.

Solution:

$\Rightarrow$ Slope of OP $\in$ (slope of the line y = 2x, slope of the line y = 3x)

$\Rightarrow$ $\alpha$ $\in$ (2, 3):

Key concept 3: First identity the locus of point $P(\alpha ,{\alpha }^2)$ , that means curve on which P lies for different values of $\alpha$. Then find the portion of the curve lying between the acute angle region between the lines.

From the graph it is clear that $\alpha \in (2,3)$

Solution: First we will find the locus of P

x = $\alpha$ and y = $\alpha$². Now on eliminating $\alpha$ we get y = x². Hence the point P will always lie on the curve y = x².

POSITION OF TWO POINTS WITH RESPECT TO A LINE

Let the given line be ax + by + c = 0 and P(x₁, y₁), Q (x₂, y₂) be two given points. Now

if $-\frac{a{x}_1+b{y}_1+c}{a{x}_2+b{y}_2+c}>0$

$\Rightarrow \frac{a{x}_1+b{y}_1+c}{a{x}_2+b{y}_2+c}<0$

$\Rightarrow$ both ax₁ + by₁ + c and ax₂ + by₂ + c are of opposite sign, then the points P and Q will lie on the opposite side of the line ax + by + c = 0 and,

if $-\frac{a{x}_1+b{y}_1+c}{a{x}_2+b{y}_2+c}<0$

$\Rightarrow \frac{a{x}_1+b{y}_1+c}{a{x}_2+b{y}_2+c}>0$

$\Rightarrow$ both ax₁ + by₁ + c and ax₂ + by₂ + c are of the same sign, then the points P and Q will lie on the same side of the given line.

Illustration 2: Find the range of $\alpha$ such that the point ($\alpha$², 2$\alpha$) and (2, 6) lies on the same side of the line x + y –1 = 0.

Key concept: Sign of x + y –1 should be same for both the points.

Solution: For (2, 6) x + y – 1 = 2 + 6 –1 is + ve. Hence $\alpha$² + 2$\alpha$ –1 > 0

$\Rightarrow$ ($\alpha$ + 1)² – 2 > 0

$\Rightarrow \text{ ( }\alpha \text{ + 1 + }\sqrt{\text{2}}\text{)}\text{ ( a + 1 - }\sqrt{\text{2}}\text{)}\text{ > 0}$

$\alpha \in (-\infty ,-1-\sqrt{2})\cup (\sqrt{2}-1,\infty )$

EQUATION OF THE STRAIGHT LINE PASSING THROUGH A POINT AND INCLINED TO A GIVE LINE

The equations of the lines through the point (x₁, y₁) and making equal angles $\alpha$ with the given line are y – y₁ = m_A(x – x₁), y – y₁ = m_B(x - x₁).

PERPENDICULAR DISTANCE OF A POINT FROM A LINE

The length of the perpendicular from

P(x₁, y₁) on ax + by + c = 0 is $\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$.

TO FIND THE IMAGE OF A POINT IN A LINE

Let the given line be ax + by + c = 0 and we have to find the image of the point P($\alpha$, $\beta$).

Working rule:

Step 1: Find the equation of line PP' (where P' is image of P). Slope of PP' will be b/a, because PP' is perpendicular to given line. Hence equation of PP' is $y-\beta =b/a(x-\alpha )$

Step 2: Find the point of intersection of line PP' with the given line and let it be R(p, q). Clearly R is the mid point of PP'.

Step 3: Now $\frac{\alpha +{\alpha }^{\prime }}{2}=p\Rightarrow {\alpha }^{\prime }=2p-\alpha$

Also $\frac{\beta +{\beta }^{\prime }}{2}=q\Rightarrow {\beta }^{\prime }=2q-\beta$

Hence the image is (2p – $\alpha$, 2q – $\beta$)

Illustration 3: Find the image of (5, 7) about line x –y + 6 = 0

Solution: slope of line PP' = –1

Hence equation of PP'

y – 7 = – 1 (x – 5) $\Rightarrow$ x + y – 12 = 0

Point of intersect of x + y – 12 = 0 with x – y + 6 = 0 is (3, 9).

Hence $\frac{{\alpha }^{\prime }+5}{2}=3and\frac{{\beta }^{\prime }+7}{2}=9$

$\frac{m-m_1}{1+m_{1}m},\text{ i}\text{.e}\text{., 1=}\pm \frac{\text{m-1/2}}{1+\frac{1}{2}m}$

CO-LINEARITY OF THREE POINTS

Following methods can be used to prove that three given points P(x₁, y₁), Q(x₂, y₂), R(x₃, y₃) are collinear.

Method 1: If P, Q and R are collinear, then area of triangle formed by these three points should be equal to zero

\Rightarrow \,\dfrac{1}{2}\,\left| \begin{gathered} {{\text{x}}_{\text{1}}}\,\,\,\,\,\,\,{{\text{y}}_1}\,\,\,\,1 \hfill \\ {{\text{x}}_{\text{2}}}\,\,\,\,\,\,\,{{\text{y}}_2}\,\,\,\,1 \hfill \\ {{\text{x}}_{\text{3}}}\,\,\,\,\,\,\,{{\text{y}}_3}\,\,\,\,1 \hfill \\ \end{gathered} \right| = 0

Method 2: Any one point of the given three points should divide other two points either internally or externally.

$\Rightarrow {\text{x}}_{\text{3}}\text{=}\frac{\text{m}{\text{x}}_{\text{2}}\text{+n}{\text{x}}_{\text{1}}}{\text{m+n}}\text{and}{\text{y}}_{\text{3}}\text{=}\frac{\text{m}{\text{y}}_{\text{2}}\text{+n}{\text{y}}_{\text{1}}}{\text{m+n}}$ for some real m and n.

Method 3: Slope of PQ = Slope of QR

$\frac{{\text{y}}_{\text{2}}\text{-}{\text{y}}_{\text{1}}}{{\text{x}}_{\text{2}}\text{-}{\text{x}}_{\text{1}}}\text{=}\frac{{\text{y}}_{\text{3}}\text{-}{\text{y}}_{\text{2}}}{{\text{x}}_{\text{3}}\text{-}{\text{x}}_{\text{2}}}$.