Q.The vertices of $\triangle PQR$ are $P(1, 0)$, $Q(5, 4)$ and $R(-1, 4)$.

Find the slope of the given line, then take the negative reciprocal for the perpendicular slope.

Rewrite $3x-4y+10=0$ in slope-intercept form:

$4y = 3x+10 \implies y = \frac{3}{4}x + \frac{10}{4}$

So the slope of the given line is $m_1 = \dfrac34$.

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Two lines with slopes $m_1$ and $m_2$ are perpendicular if and only if $m_1 m_2 = -1$.

If a line $L_1$ makes angle $\theta$ with the x-axis, its slope is $m_1=\tan\theta$. A line perpendicular to it makes angle $\theta+90°$, with slope: …

Perpendicular lines have slopes whose product is $-1$; solving gives $a=8$.

Rewrite each line in slope form $y=mx+c$:

• $3x-4y+7=0 \Rightarrow y=\dfrac{3}{4}x+\dfrac{7}{4}$, slope $m_1=\dfrac34$
• $ax+6y+1=0 \Rightarrow y=-\dfrac{a}{6}x-\dfrac16$, slope $m_2=-\dfrac{a}{6}$ …

Find slope of AS, then take the negative reciprocal.

A$(3,1)$, S$(-3,3)$.

Slope of AS $=\dfrac{3-1}{-3-3}=\dfrac{2}{-6}=-\dfrac13$.

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The standard perpendicularity condition for two lines is $m_1 m_2 = -1$.

For two non-vertical lines with slopes $m_1$ and $m_2$, they are perpendicular if and only if $m_1 \cdot m_2 = -1$ (one slope is the negative reciprocal of the other).

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Find the first line's slope, use $m_1m_2=-1$ to get the second slope, then solve for $x$.

Slope of the line through $(-2,6)$ and $(4,8)$:

$m_1=\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}$

For two lines to be perpendicular, $m_1 m_2 = -1$, so $m_2 = -3$.

Slope of the line through $(8,12)$ and $(x,24)$:

$m_2=\frac{24-12}{x-8}=\frac{12}{x-8}$ …

The orthocentre is $(3,4)$.

The sides $x=3$ (vertical) and $y=4$ (horizontal) are perpendicular and meet at $(3,4)$, so the triangle has a right angle at that vertex.

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Two lines are perpendicular when the product of their slopes is $-1$; find the slope of $3x+5y=8$ and solve for $m$.

The line $y = mx+5$ has slope $m_1 = m$.

Rewrite $3x+5y=8$ in slope form: $5y = -3x+8 \Rightarrow y = -\dfrac{3}{5}x + \dfrac{8}{5}$, so its slope is $m_2 = -\dfrac{3}{5}$.

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