If the sides of a triangle are given by y = m | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The lines are $L_1: y = mx + a$,$L_2: y = nx + b$,and $L_3: x = 0$ (the $y$-axis).The intersection of $L_1$ and $L_3$ is $A = (0, a)$.The intersec

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The lines are $L_1: y = mx + a$,$L_2: y = nx + b$,and $L_3: x = 0$ (the $y$-axis).The intersection of $L_1$ and $L_3$ is $A = (0, a)$.The intersection of $L_2$ and $L_3$ is $B = (0, b)$.The base of the triangle along the $y$-axis is the distance between $A$ and $B$,which is $|a - b|$.To find the third vertex $C$,we solve $mx + a = nx + b$:$(m - n)x = b - a \Rightarrow x = \frac{b - a}{m - n} = \frac{a - b}{n - m}$.The height of the triangle is the absolute value of the $x$-coordinate of $C$,which is $h = |\frac{a - b}{n - m}|$.The area of the triangle is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes |a - b| 	imes |\frac{a - b}{n - m}| = \frac{(a - b)^2}{2|m - n|}$.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

If the sides of a triangle are given by $y = mx + a$,$y = nx + b$,and $x = 0$,then its area is:

Similar Questions

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

$(0, -1)$ and $(0, 3)$ are two opposite vertices of a square. The other two vertices are

Given two points $A(3, 4)$ and $B(5, -2)$. If $PA = PB$ and the area of $\Delta PAB = 10$ square units,find the coordinates of point $P$.

Two mutually perpendicular straight lines through the origin form an isosceles triangle with the line $2x + y = 5$. Then the area of the triangle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module