The extremities of the base of an isosceles triangle A B C are the points A (2,0) and B (0,1). If the equation of the side A C is x=2 then the slope of the side B C is

Question

The extremities of the base of an isosceles triangle A B C are the points A (2,0) and B (0,1). If the equation of the side A C is x=2 then the slope of the side B C is (A) (3/4) (B) (4/3) (C) (3/2) (D) √3

Tardigrade Admin · Accepted Answer

AB = BC ∴ √ k 2=√4+( k -1)2 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/a440607b33b6748b70ccfe2c49fa43ed-.png" /> k 2=4+ k 2-2 k +1 ∴ k =(5/2) ∴ Slope of BC =((5/2)-1/2)=(5-2/4)=(3/4)

Q. The extremities of the base of an isosceles triangle $A BC$ are the points $A (2, 0)$ and $B (0, 1)$ . If the equation of the side $A C$ is $x = 2$ then the slope of the side $BC$ is

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{3}{2}$

$3$

$A B = BC$
$∴ k^{2} = 4 + (k - 1)^{2}$

$k^{2} = 4 + k^{2} - 2 k + 1$
$∴ k = \frac{5}{2}$
$∴$ Slope of $BC = \frac{\frac{5}{2} - 1}{2} = \frac{5 - 2}{4} = \frac{3}{4}$

Q. The extremities of the base of an isosceles triangle ABC are the points A(2,0) and B(0,1). If the equation of the side AC is x=2 then the slope of the side BC is

43​

34​

23​

3​

AB=BC ∴k2​=4+(k−1)2​ k2=4+k2−2k+1 ∴k=25​ ∴ Slope of BC=225​−1​=45−2​=43​

Q. The extremities of the base of an isosceles triangle $A BC$ are the points $A (2, 0)$ and $B (0, 1)$ . If the equation of the side $A C$ is $x = 2$ then the slope of the side $BC$ is

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{3}{2}$

$3$

$A B = BC$
$∴ k^{2} = 4 + (k - 1)^{2}$

$k^{2} = 4 + k^{2} - 2 k + 1$
$∴ k = \frac{5}{2}$
$∴$ Slope of $BC = \frac{\frac{5}{2} - 1}{2} = \frac{5 - 2}{4} = \frac{3}{4}$