# ML Aggarwal Solutions Class 10 Mathematics Solutions for Equation of Straight Line Exercise 12.2 in Chapter 12 - Equation of Straight Line

Question 40 Equation of Straight Line Exercise 12.2

A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC.

Find the equation of:  the altitude of the triangle through B.

Solution:

The slope of a line that is perpendicular to another line is equal to the negative reciprocal of that line's slope.

Given, A (2, – 4), B (3, 3) and C (– 1, 5) are the vertices of triangle ABC

D is the mid-point of BC

So, the co-ordinates of D will be

((3 – 1)/2, (3 + 5)/2) = (2/2, 8/2) = (1, 4)

Now,

The slope of AC (m1) = (5 + 4)/ (-1 – 2) = 9/-3 = -3

Let the slope of BE be m2

Then, m1 x m2 = -1

-3 x m2 = -1

m2 = 1/3

so, the equation of BE will be

y – 3 = 1/3 (x – 3)

3y – 9 = x – 3

x – 3y + 6 = 0

Thus, the required line equation is x – 3y + 6 = 0.

Related Questions

Lido

Courses

Teachers

Book a Demo with us

Syllabus

Maths
CBSE
Maths
ICSE
Science
CBSE

Science
ICSE
English
CBSE
English
ICSE
Coding

Terms & Policies

Selina Question Bank

Maths
Physics
Biology

Allied Question Bank

Chemistry
Connect with us on social media!