If the vertex opposite to the hypotenuse of a right-angled triangle lies on the straight line 2x+y-10=0 and the two other vertices are (2 , - 3) and (4 , 1) , then the area of the triangle (in sq. units) is equal to

Question

If the vertex opposite to the hypotenuse of a right-angled triangle lies on the straight line 2x+y-10=0 and the two other vertices are (2 , - 3) and (4 , 1) , then the area of the triangle (in sq. units) is equal to (A) √10 (B) 3 (C) (33/5) (D) 11

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-0npydmznfmrn.png" /> m1m2=-1⇒ (9 - 2 a/a - 4)× (13 - 2 a/a - 2)=-1 ⇒ 5a2-50a+125=0 So, a=5 and B is (5 , 0) Hence, required area =(1/2)AB× AC=(1/2)√2× 3√2=3 sq. units

Q. If the vertex opposite to the hypotenuse of a right-angled triangle lies on the straight line $2 x + y - 10 = 0$ and the two other vertices are $(2, - 3)$ and $(4, 1)$ , then the area of the triangle (in sq. units) is equal to

$10$

$3$

$\frac{33}{5}$

$11$

$m_{1} m_{2} = - 1 \Rightarrow \frac{9 - 2 a}{a - 4} \times \frac{13 - 2 a}{a - 2} = - 1$
$\Rightarrow 5 a^{2} - 50 a + 125 = 0$
So, $a = 5$ and $B$ is $(5, 0)$
Hence, required area $= \frac{1}{2} A B \times A C = \frac{1}{2} 2 \times 32 = 3$ sq. units

Q. If the vertex opposite to the hypotenuse of a right-angled triangle lies on the straight line 2x+y−10=0 and the two other vertices are (2,−3) and (4,1) , then the area of the triangle (in sq. units) is equal to

10​

3

533​

11

m1​m2​=−1⇒a−49−2a​×a−213−2a​=−1 ⇒5a2−50a+125=0 So, a=5 and B is (5,0) Hence, required area =21​AB×AC=21​2​×32​=3 sq. units

Q. If the vertex opposite to the hypotenuse of a right-angled triangle lies on the straight line $2 x + y - 10 = 0$ and the two other vertices are $(2, - 3)$ and $(4, 1)$ , then the area of the triangle (in sq. units) is equal to

$10$

$3$

$\frac{33}{5}$

$11$

$m_{1} m_{2} = - 1 \Rightarrow \frac{9 - 2 a}{a - 4} \times \frac{13 - 2 a}{a - 2} = - 1$
$\Rightarrow 5 a^{2} - 50 a + 125 = 0$
So, $a = 5$ and $B$ is $(5, 0)$
Hence, required area $= \frac{1}{2} A B \times A C = \frac{1}{2} 2 \times 32 = 3$ sq. units