JEE Main 2025 Apr 4 Shift 1, Mathematics Q15: Solution of Triangles

JEE Main2025Apr 4, Shift 1Mathematics

Let the three sides of a triangle are on the lines $4 x - 7 y + 10 = 0$ , $x + y = 5$ and $7 x + 4 y = 15$ . Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0$ , $y = 0$ and $x + y = 1$ is

A
5
B
$5$
C
$20$
D
20

Solution

Check perpendicularity of $4 x - 7 y + 10 = 0$ and $7 x + 4 y = 15$ : slopes are $\frac{4}{7}$ and $- \frac{7}{4}$ , whose product is $- 1$ . So these two lines meet at right angles; the triangle is right-angled at their point of intersection, and the orthocentre of a right triangle is that right-angle vertex.

Solving $4 x - 7 y + 10 = 0$ and $7 x + 4 y = 15$ simultaneously gives $x = 1$ , $y = 2$ . So the orthocentre of the first triangle is $H_{1} = (1, 2)$ .

For the second triangle (with vertices $(0, 0)$ , $(1, 0)$ , $(0, 1)$ ), the right angle is at the origin, so its orthocentre is $H_{2} = (0, 0)$ .

Distance $= 1^{2} + 2^{2} = 5$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q15: Solution of Triangles

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