Question

If sides of a triangle are y=mx+a,y=nx+b and , x=0 ,then its area is- (1) (a-b)^(2)/(2(m-n))," (2) "(1)/(2)((a-b)^(2))/((m+n)), (3) "(a+b)^(2))/(2(m-n))," (4) none ​

Answer 1

Answer:

Option (1) $\frac{(a-b)^2}{2(m-n)}$

Step-by-step explanation:

Given lines

$L_1: y=mx+a$

$L_1: y=nx+b$

Line $L_1$ intersects y-axis at A = (0, a)

Line $L_1$ intersects y-axis at B = (0, b)

Solving the above two equations

$mx+a=nx+b$

$\implies nx-mx=a-b$

$\implies x=\frac{a-b}{n-m}$

Thus from $L_1$

$y=m\frac{a-b}{n-m}+a$

$\implies y=\frac{ma-mb+na-ma}{n-m}$

$\implies y=\frac{na-mb}{n-m}$

Thus the coordinates of intersection of the lines is C = $(\frac{a-b}{n-m},\frac{na-mb}{n-m})$

Distance between A and B

$d=\sqrt{(0-0)^2+(a-b)^2}$

$\implies d=\sqrt{(a-b)^2}$

$\implies d=a-b$

Length of perpendicular from C on the y-axis

$p=\frac{a-b}{n-m}$

Therefore, area of the triangle formed by the lines $L_1, L_2$ and y-axis

$A=|\frac{1}{2}dp|$

$\implies A=|\frac{1}{2}\times (a-b)\times\frac{a-b}{n-m}|$

$\implies A=|\frac{(a-b)^2}{2(m-n)}|$

Therefore, option (1) is correct.

Hope this helps.

Answer 2

Step-by-step explanation:

use the formula to have your answer

area =

$\frac{1}{2} | ({c1 - c2})^{2} \div (m1 - m2)|$