Rearrange to make $y$ the subject first — the gradient is the coefficient of $x$ only once $y$ is alone, never before.

Find the gradient of $2x+5y=15$: $5y=-2x+15 \Rightarrow y = -\frac{2}{5}x+3$, so $m=-\frac{2}{5}$ — not $2$ or $5$, the raw coefficients from the original form.

Parallel lines share the same gradient: $m_2 = m_1$. The fastest route to the new equation is the point–gradient form $y-y_1=m(x-x_1)$ — substitute the gradient and the given point, then rearrange.

Line parallel to $y=3x-5$ through $(2,7)$: $m=3$ (unchanged), so $y-7=3(x-2) \Rightarrow y=3x+1$.

Same idea when the given line is in $ax+by=c$ form: rearrange first, don't guess the gradient from the raw coefficients.

Line parallel to $2x+5y=15$ through $(10,-1)$: rearranged, $y=-\frac25x+3$, so $m_1=-\frac25$ (not $2$). Parallel line: $y-(-1)=-\frac25(x-10) \Rightarrow y+1=-\frac25x+4 \Rightarrow y=-\frac25x+3$. If an integer-coefficient form is required, multiply by 5: $2x+5y=15$.

Perpendicular gradients are negative reciprocals: $m_2=-\dfrac{1}{m_1}$, equivalently $m_1 \times m_2 = -1$. In practice: flip the fraction, then switch the sign.

Perpendicular to $y=2x+5$ through $(3,7)$: $m_1=2 \Rightarrow m_2=-\frac12$. Then $y-7=-\frac12(x-3) \Rightarrow y=-\frac12x+\frac{17}{2}$.

If the original line is given as $ax+by=c$, rearrange to find $m_1$ before flipping — the same trap as with parallel lines.

Perpendicular to $4x-3y=12$ through $(8,0)$: rearrange: $y=\frac43x-4 \Rightarrow m_1=\frac43 \Rightarrow m_2=-\frac34$. Then $y-0=-\frac34(x-8) \Rightarrow y=-\frac34x+6$.

The reciprocal rule breaks down here since $m=0$ has no reciprocal — reason from the picture instead. A horizontal line ($y=c$) and a vertical line ($x=k$) are always perpendicular to each other, and parallel lines in either family just share that same fixed coordinate.

• Parallel to $y=4$ through $(1,6)$: still horizontal, so $y=6$.
• Perpendicular to $y=4$ through $(1,6)$: must be vertical, so $x=1$.

"Bisector" means it passes through the midpoint; "perpendicular" means its gradient is the negative reciprocal of the segment's gradient. Combine both:

Perpendicular bisector of $A(2,5)$ and $B(10,1)$:

1. Gradient of $AB$: $m_1=\dfrac{1-5}{10-2}=-\dfrac12$.
2. Perpendicular gradient: $m_2 = 2$.
3. Midpoint of $AB$: $\left(\dfrac{2+10}{2},\dfrac{5+1}{2}\right)=(6,3)$.
4. Equation through $(6,3)$ with gradient $2$: $y-3=2(x-6) \Rightarrow y=2x-9$.

Find the equation of the line perpendicular to $y=\frac{1}{5}x+2$ that passes through the point $(-5, 4)$.

1. $m_1 = \frac15 \Rightarrow m_2 = -5$ (flip to $\frac51$, change sign).
2. $y - 4 = -5(x-(-5)) = -5(x+5)$.
3. Expand: $y-4=-5x-25 \Rightarrow y=-5x-21$.

Always double-check by confirming $m_1 \times m_2 = \frac15 \times (-5) = -1$ — a 10-second sanity check that catches the most common error on the whole topic.