Question

Find the equation of line passes through the point (-3,8) and cuts off positive intercepts on the axes whose sum is 7

Answer 1

Hey there

refer to attachment

Answer 2

Answer:

4x+3y-12=0

Step-by-step explanation:

Given sum of intercepts at the axes is 7.

i.e., a+b=7

or b=7-a

Also let the point of contact of the line on the y-axis be A, and on the x-axis be B.

So the coordinates of A and B are A=(0,b) B=(a,0)

Since b=7-a, A=(0,7-a),B=(a,0)

Now we have two points to find the equation of the line AB.

We know that the equation of a line joining two points (x1, y1) & (x2, y2) is given by

(y-y1)/(y2-y1 )=(x-x1)/(x2-x1 )

So, equation of AB with points (0,7-a) & (a,0) is given by

(y-(7-a))/(0-(7-a))=(x-0)/(a-0)

(y-7+a)/(-7+a)=x/a

a(y-7+a)=x(-7+a)

ya-7a+a^2=-7x+ax

Let this be equation 1.

Since the line passes through the point (-3,8), it should satisfy equation 1

Substituting (-3,8) in equation 1 we get

ya-7a+a^2=-7x+ax

8a-7a+a^2=-7(-3)+a(-3)

a+a^2=21-3a

4a+a^2-21=0

a^2+4a-21=0

This is a quadratic in a Solving the quadratic equation we get

a^2+7a-3a-21=0

a(a+7)-3(a+7)=0

(a+7)(a-3)=0

a=-7 or a=3

a≠-7 as the line cuts the positive intercept, so a=3

Put a=3 in equation 1

ya-7a+a^2=-7x+ax

y(3)-7(3)+3^2=-7x+(3)x

3y-21+9=-7x+3x

3y-12=-4x

4x+3y-12=0

Therefore, the equation is 4x+3y-12=0