Question bank — Distance from a point to a line

TRUE. Any slanted path is the hypotenuse of a right triangle whose perpendicular leg is that shortest path, and a hypotenuse always exceeds a leg.

FALSE. Numerator and denominator both scale by $5$, so the $5$ cancels — the distance depends on the line, not on the particular numbers you wrote it with.

TRUE. Then $a{x}_1+b{y}_1+c=0$, the numerator is $0$, and $d=0$ — exactly the distance you'd expect.

FALSE. If $a=b=0$ the equation becomes $c=0$, which is not a line at all. A real line always has at least one of $a,b$ nonzero, so the denominator is positive.

TRUE. Parallel lines are equidistant everywhere, so any point on one gives the same perpendicular distance to the other.

FALSE as written. The formula with $|\cdots |$ gives only magnitude; drop the bars and then the sign of $a{x}_1+b{y}_1+c$ encodes the side (positive on the normal's side, negative on the other).

TRUE. Distance is a magnitude; a point and its mirror image across the line share the same $d$ but opposite signed values of $a{x}_1+b{y}_1+c$.

FALSE. The coefficients must match first. Halve the second line to $2x+y-3=0$; then $d=\frac{|-1-(-3)|}{\sqrt{5}}=\frac{2}{\sqrt{5}}$.

The absolute value is missing. A distance cannot be negative, so the numerator must be $|a{x}_1+b{y}_1+c|$.

The square root is gone. The denominator is the length of the normal vector $(a,b)$, which is $\sqrt{a^2+b^2}$, not $a^2+b^2$.

The line isn't in $ax+by+c=0$ form. Rewrite as $2x-y+3=0$, giving $a=2,b=-1,c=3$ — note $b$ is $-1$, not $3$.

The constant $c=-10$ must be included: $\frac{|9+16-10|}{5}=3$. Dropping $c$ measures distance to the parallel line through the origin, not to $L$.

Two mistakes: the intercept was dropped (line is $y-1=\frac{4}{3}(x-1)$), and the slope form wasn't converted. Correct standard form is $4x-3y-1=0$.

The denominator is $\sqrt{1^2+(-2{)}^2}=\sqrt{5}$, not $1+(-2)$. Coefficients get squared and rooted, never just added.

Only the $x$ and $y$ coefficients form the normal vector: $\vec{n}=(3,4)$. The constant $c$ shifts the line but has no direction.

Because a slanted route is the hypotenuse of a right triangle whose vertical leg is the perpendicular drop; the hypotenuse is always longer, so perpendicular wins.

It is the length of the normal vector $(a,b)$. Dividing by it converts a raw dot-product ($a{x}_1+b{y}_1+c$) into an actual distance measured in the same units as the axes.

That expression is the dot product of the normal with $P$'s position (plus $c$); it measures how far along the normal direction $P$ sits, which — once scaled by $1/\sqrt{a^2+b^2}$ — is exactly the perpendicular offset.

The formula reads off $a,b,c$ directly; slope-intercept form hides them and would give wrong coefficients (and a bogus normal vector) if plugged in as-is.

Multiplying by $k$ scales numerator and denominator equally, so $k$ cancels — the geometry of the line is the same, only its algebraic label changed.

Distance needs pure magnitude, so bars stay. Side-detection needs the sign of $a{x}_1+b{y}_1+c$ to distinguish the two half-planes, so bars come off.

Exactly $0$, since $a{x}_1+b{y}_1+c=0$ makes the numerator vanish — consistent with $P$ already being on the road.

Here $a=1,b=0,c=-5$, so $d=\frac{|x_1-5|}{\sqrt{1}}=|x_1-5|$ — just the horizontal gap, as geometry demands.

With $a=0,b=1,c=2$, $d=\frac{|y_1+2|}{1}=|y_1+2|$ — the pure vertical gap, independent of $x_1$.

No. $c=0$ just means the line passes through the origin; the formula handles any sign of $c$ identically because everything sits inside the absolute value.

Yes, proportional coefficients mean parallel lines; but you must first rescale one so both share identical $(a,b)$ before applying $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$.

No. The inflated numerator is exactly matched by the inflated denominator, so $d$ is unaffected — a reassuring sign the formula measures true geometry.