Question

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The vertices of a ∆ABC are A(3,8) B(-1,2) and C(6,-6).
Find =>
1) Slope of BC
2) Equation of a line perpendicular to BC and passing through A.​

Answer 1

Solution :


Slope of a line = $\sf \: \frac{dy}{dx}$

Slope of BC =

$= \frac{ - 6 - 2}{6 - ( - 1)} \\ \\ = \frac{ - 8}{7}$

2) For two lines to be perpendicular, Product of their slopes need to be - 1.

Now, We already know Slope of BC ( - 8/7).

So, Let Slope of the other line be m.

By the relationship stated before ,

$m\times \frac{ - 8}{7} = 1 \\ \\ m = \frac{ - 7}{8}$

We have the value of Slope,

So, Equation of the line will be y = mx + c

y = mx + c

y = - 7/8x + c

8y = - 7x + c

As the question states, The perpendicular passes through A ( 3,8)

8y = - 7x + c

8(8) = - 7(3) + c

64 + 21 = c

85 = c.

Therefore, The equation of line perpendicular to BC is 7x + 8y = 85.

Answer 2

The vertices of a ∆ABC are A(3,8) B(-1,2) and C(6,-6).

1) Let B(-1,2)=(x1,y1)

C(6,-6) = (x2,y2)

Slope of BC = (y2-y1)/(x2-x1)

= (-6-2)/[6-(-1)]

= -8/7

2) Slope of a line perpendicular

to BC(m) = (-7/8)

It is passing through the point

(x1,y1) = A(3,8)

Now ,

Required equation:

y-y1 = m(x-x1)

=> y -8 = (-7/8)(x-3)

=> 8(y-8) = -7(x-3)

=> 7(x-3)+8(y-8)=0

=> 7x-21+8y-64=0

=> 7x+8y-85=0

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