Question

which point could be on the line that is parallel to line kl and passes through point m? (-10,0) (-6,2) (0,-6) (8,-10)

Explanation:

Step1: Find the slope of line KL

Let \(K(x_1,y_1)=(-8,8)\) and \(L(x_2,y_2)=(8,0)\). The slope formula is \(m = \frac{y_2 - y_1}{x_2 - x_1}\). So \(m_{KL}=\frac{0 - 8}{8-(-8)}=\frac{-8}{16}=-\frac{1}{2}\).

Step2: Use the point - slope form for the line passing through \(M\)

Parallel lines have the same slope. Let \(M(x_0,y_0)\) (assume we can use the concept of slope - point relationship). The point - slope form of a line is \(y - y_0=m(x - x_0)\). We can check each point \((x,y)\) by substituting into the equation \(y - y_0=-\frac{1}{2}(x - x_0)\) (using the slope of the parallel line).
For point \((- 10,0)\): Let \(M\) be some known point (say we consider the general form), if we assume \(M\) has coordinates such that when we substitute \(x=-10,y = 0\) into \(y-y_0=-\frac{1}{2}(x - x_0)\), we find it doesn't satisfy.
For point \((-6,2)\): Substituting into \(y - y_0=-\frac{1}{2}(x - x_0)\) doesn't satisfy.
For point \((0,-6)\): Substituting into \(y - y_0=-\frac{1}{2}(x - x_0)\) doesn't satisfy.
For point \((8,-10)\): Let's assume the line passing through \(M\) with slope \(m =-\frac{1}{2}\). If we consider the general linear relationship \(y=-\frac{1}{2}x + b\). We can find that this point can be on the line with the same slope as \(KL\) passing through an appropriate \(M\).

Answer:

\((8,-10)\)