Suppose that the points (h, k),(1, 2),and (-3 | vedclass

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(A) Slope of line $l_1 = \frac{4-2}{-3-1} = \frac{2}{-4} = -\frac{1}{2}$.Since $(h, k)$ lies on $l_1$,the slope between $(h, k)$ and $(1, 2)$ must be

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$\frac{1}{3}$

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(A) Slope of line $l_1 = \frac{4-2}{-3-1} = \frac{2}{-4} = -\frac{1}{2}$.Since $(h, k)$ lies on $l_1$,the slope between $(h, k)$ and $(1, 2)$ must be $-\frac{1}{2}$:$\frac{k-2}{h-1} = -\frac{1}{2}$ $\Rightarrow 2k-4 = -h+1$ $\Rightarrow h+2k = 5$ ... $(i)$.Since $l_2$ passes through $(h, k)$ and $(4, 3)$ and is perpendicular to $l_1$,its slope is $m_2 = -\frac{1}{m_1} = -\frac{1}{-1/2} = 2$.Thus,$\frac{3-k}{4-h} = 2$ $\Rightarrow 3-k = 8-2h$ $\Rightarrow 2h-k = 5$ ... $(ii)$.Multiplying $(ii)$ by $2$: $4h-2k = 10$ ... $(iii)$.Adding $(i)$ and $(iii)$: $(h+2k) + (4h-2k) = 5+10$ $\Rightarrow 5h = 15$ $\Rightarrow h = 3$.Substituting $h=3$ into $(i)$: $3+2k = 5$ $\Rightarrow 2k = 2$ $\Rightarrow k = 1$.Therefore,$\frac{k}{h} = \frac{1}{3}$.

Suppose that the points $(h, k)$,$(1, 2)$,and $(-3, 4)$ lie on the line $l_1$. If a line $l_2$ passing through the points $(h, k)$ and $(4, 3)$ is perpendicular to $l_1$,then $\left(\frac{k}{h}\right)$ equals

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