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

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(B) The slope of line $L_1$ passing through $(1, 2)$ and $(-3, 4)$ is $m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$.Since $(h, k)$ lies o

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$\frac{1}{3}$

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(B) The slope of line $L_1$ passing through $(1, 2)$ and $(-3, 4)$ is $m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2}$.Since $(h, k)$ lies on $L_1$,the slope of the line segment joining $(h, k)$ and $(1, 2)$ must also be $-\frac{1}{2}$.Thus,$\frac{k - 2}{h - 1} = -\frac{1}{2}$ $\Rightarrow 2k - 4 = -h + 1$ $\Rightarrow h + 2k = 5$.Line $L_2$ passes through $(h, k)$ and $(4, 3)$,so its slope is $m_2 = \frac{3 - k}{4 - h}$.Since $L_2 \perp L_1$,$m_1 	imes m_2 = -1$ $\Rightarrow -\frac{1}{2} 	imes \frac{3 - k}{4 - h} = -1$ $\Rightarrow \frac{3 - k}{4 - h} = 2$.$3 - k = 8 - 2h \Rightarrow 2h - k = 5$.We have the system of equations:$1) h + 2k = 5$$2) 2h - k = 5$Multiplying equation $(2)$ by $2$,we get $4h - 2k = 10$.Adding this to equation $(1)$: $(h + 2k) + (4h - 2k) = 5 + 10$ $\Rightarrow 5h = 15$ $\Rightarrow h = 3$.Substituting $h = 3$ into $2h - k = 5$: $2(3) - k = 5$ $\Rightarrow 6 - k = 5$ $\Rightarrow k = 1$.Therefore,$\frac{k}{h} = \frac{1}{3}$.

Column $I$	Column $II$
$(a)$ Proliferative Phase	$(i)$ Breakdown of endometrium
$(b)$ Secretory Phase	$(ii)$ Follicular phase
$(c)$ Menstruation	$(iii)$ Luteal Phase

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 $\frac{k}{h}$ equals

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$(a)$ Proliferative Phase $(i)$ Breakdown of endometrium

$(b)$ Secretory Phase $(ii)$ Follicular phase

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