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I have no answers for these questions.. please help me!

boujeemath

New member
Jul 5, 2020
1
I'm working on my geometry during winter break and I'm solving questions from the cambridge book but they don't have worked soutions :((

Screen Shot 2020-07-06 at 12.09.59 pm.png
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,642
Hi boujeemath, welcome to MHB!

For question 22, we can let the $y$-intercept be $(0,\,a)$ and $x$-intercept be $(b,\,0)$. Since we are told $M(4,\,3)$ is the midpoint of the intercepts, we have

$\left(\dfrac{b}{2},\,\dfrac{a}{2}\right)= (4,\,3)$

That gives us $b=8$ and $a=6$.

Now, you have found the intercepts, can you proceed to find the equation of the line?
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,642
To follow up, once we have the $x$- and $y$ intercepts, we know the gradient, $m=-\dfrac{y\text{-intercept}}{x\text{-intercept}}=-\dfrac{6}{8}=-\dfrac{3}{4}$.

Therefore, taking $m=-\dfrac{3}{4}$ and the point $(8,\,0)$, we find that the point=gradient form of the straight line is:

$y-0=\dfrac{3}{4}\left(x-8\right)$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
428
Problem 23 asks you to, first, write the general equation for a line through the point (-1, 2) with gradient m. I don't know what form you have learned as the general equation for a line through (a, b) with gradient m. One is [tex]\frac{y- b}{x- a}= m[/tex]. Another is y- b= m(x- a) which you can get by multiplying both sides of the first by x- a. Here a= -1 and b= 2. so y- 2= m(x+ 1) or y= m(x+1)+ 2 or y= mx+ m+ 2..

The problem then asks you to find the equation of the specific line when (-1, 2) is "the midpoint of the intercepts with the x and y axes". y- 2= m(x+ 1) "intercepts" the y axis where x= 0. That is y- 2= m(1) so y= m+ 2. The y-intercept is (0, m+2). It intercepts the x- axis where y= m(x+ 1)+ 2= 0. m(x+ 1)= -2. x+1= -2/m, x= -1- 2/m. The x-intercept is (-1-2/m, 0). The midpoint is (-1/2-1/m, m/2+ 1). You want -1/2- 1/m= -1 and m/2+ 1= 2. Of course, that is asking for one value to satisfy two differential equations. In general we cannot do that but from m/2+ 1= 2 we have m/2= 1 and then m= 2. Putting that into -1/2- 1/m= -1 we get -1/2- 1/2= -1 which is a true statement! The line we want is y= 2(x+ 1)+ 2= 2x+ 4.

ChecK: The line y= 2x+ 4 has x-intercept (0, 4) and y-intercept (-2, 0). The midpoint is ((-2+ 0/2, (4+ 0)/2)= (-1, 2).