Find the perpendicular distance from Q to line MF₁.

01 / Standard#Your turn

Q=\left(\frac{3\sqrt5}5,\frac{4\sqrt5}5\right)

Official paper · jm02-2026 · 3(d) · PDF 5

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Write MF₁ in general form.

Hint 2

Apply the point-to-line distance formula.

Worked solution

Use slope −2/11 and point F₁.
$MF_1:\ 2x+11y+2\sqrt5=0$
Substitute Q and normalise the normal vector.
$d=\frac{|2(3\sqrt5/5)+11(4\sqrt5/5)+2\sqrt5|}{\sqrt{2^2+11^2}}=\frac{12\sqrt5}{5\sqrt5}=\frac{12}5$

Distance 12/5.

Checks and common pitfalls: The distance denominator is √(A²+B²), not A²+B².

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
Use slope −2/11 and point F₁.
$MF_1:\ 2x+11y+2\sqrt5=0$
Substitute Q and normalise the normal vector.
$d=\frac{|2(3\sqrt5/5)+11(4\sqrt5/5)+2\sqrt5|}{\sqrt{2^2+11^2}}=\frac{12\sqrt5}{5\sqrt5}=\frac{12}5$

Think first. Reveal a hint when the class is ready.

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