Question

find the sum of the squares of the intercepts of the line 4x-3y=12 on the axes of co ordinates

Answer 1

4x/12 -3y/12 =1
x/3 - y/4 =1
intercepts a, b = 3,4
a^2 + b^2 = (3)^2 + (4)^2
= 9 + 16 =25
a^2 + b^2 = 25 is the obtained answer

Answer 2

we have to find the sum of squares of the intercepts of the line 4x - 3y = 12 on the axes of co-ordinates.

to find x - intercept, take y = 0

i.e., 4x - 3 × 0 = 12 ⇒x = 3

To find y - intercept , take x = 0

i.e., 4 × 0 - 3y = 12 ⇒y = -4

now sum of square of intercepts = (x - intercept)² + (y - intercept)²

= (3)² + (-4)²

= 9 + 16 = 25

also read similar questions : Find equations of lines which contains

the point A(1,3) and the sum of whose

intercepts on the co-ordinate axes is zero.

https://brainly.in/question/12733138

A straight line makes on the co-ordinate axes positive intercepts whose sum is 5.If the line passes through the  point  ...

https://brainly.in/question/38936