Lines in the Plane practice questions — SAT Math

$2x+6=0\Rightarrow x=-3$, so the line crosses the $x$-axis at $(-3,0)$.

$\overrightarrow{AB}=(2,2)$, $\overrightarrow{AC}=(a-1,3)$. Collinear ⇒ ${\displaystyle \frac{a-1}{2}}={\displaystyle \frac{3}{2}}\Rightarrow a-1=3\Rightarrow a=4$.

$m_1\cdot m_2=-1\Rightarrow m_2=-{\displaystyle \frac{1}{2}}$.

Both lines have slope $2$ but different $y$-intercepts ($3$ and $-1$).

Same slope, different intercept ⇒ parallel and distinct (never meet).

With $(x_0,y_0)=(0,0)$ and the line $3x+4y-10=0$:

$d={\displaystyle \frac{\mathrm{\mid }3(0)+4(0)-10\mathrm{\mid }}{\sqrt{3^2+4^2}}}={\displaystyle \frac{\mathrm{\mid }-10\mathrm{\mid }}{\sqrt{25}}}={\displaystyle \frac{10}{5}}=2$.

Slope $=-3$. Line through $(1,4)$: $y-4=-3(x-1)\Rightarrow y=-3x+7$.

$d={\displaystyle \frac{\mathrm{\mid }3\cdot 2-4\cdot 3+5\mathrm{\mid }}{\sqrt{9+16}}}={\displaystyle \frac{\mathrm{\mid }-1\mathrm{\mid }}{5}}={\displaystyle \frac{1}{5}}$.

The constant term in slope-intercept form is the $y$-intercept: $b=5$.

Slope $=2$. The line through $(0,1)$: $y-1=2(x-0)\Rightarrow y=2x+1$.

The line is in slope-intercept form $y=mx+b$ with $m=-3$ and $b=5$.

So the slope is $-3$.