Through the point $A(4,5)$ a line is drawn.

Question

Through the point $A(4,5)$ a line is drawn inclined at $45°$ with the $+ve$ X - axis. It meets $x+y=6$ at the point $B$. Find the equation of $AB$.

My solution..

Equation of $AB$

$$(y-y_1)=m(x-x_1)$$ $$(y-5)=1(x-4)$$ $$x-y+1=0$$.

But the answer in my book is $3x-y=7$.

Can anyone tell me where I made mistake?.

Answer 1

You only figured out the equation of the line through A, at an angle of $45^\circ$ with the $x$-axis, what you need to do is:

Find it's point of intersection with the other line $x+y=6$,you will get point $B (\frac{5}{2},\frac{7}{2}$) , and then write the equation of AB in the form of $$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$$ Where A is $(x_1,y_1)$ and B is $(x_2,y_2)$.

Answer 2

The equation of the first line is: $y - 5 = \tan(45^{\circ})(x - 4)= 1(x-4) = x-4\Rightarrow y = 5+x-4 = x+1$. Thus the intersection is found by: $x + 1 = 6- x \Rightarrow x = \dfrac{5}{2}\Rightarrow y = 6 - \dfrac{5}{2} = \dfrac{7}{2}\Rightarrow B = (\frac{5}{2}, \frac{7}{2})$. Can you take it from here?

Answer 3

Your answer is correct. You don't even need to figure out where point B is. You know the original line goes through point A and point B, and the equation for that line is as you wrote: y=x+1.