Type 1: Solve for parameters using distance

1. The distance from point \(A(2,a)\) to the line \(3x-4y+2=0\) is 2. Find the value of a.
2. The distance from point \((m,1)\) to the line \(x+2y-5=0\) equals its distance to the y-axis. Find m.

Type 2: Parallel lines distance & unknown line equation

3. Find the equation of the straight line parallel to \(2x-3y+6=0\) and at a distance \(\sqrt{13}\) from it.
4. Given two parallel lines \(x+ay-1=0\) and \(2x+2y+3=0\), find a and the distance between the two lines.

Type 3: Minimum value problems (geometric meaning of distance)

5. Point P lies on the line \(x+y-4=0\). Find the minimum distance from the origin O to point P.
6. Real numbers \(x,y\) satisfy \(x+y-3=0\). Find the minimum value of \(\sqrt{x^2+y^2}\).

Type 4: Angle bisector & locus / straight line with fixed distance

7. Find the locus equation of points equidistant from lines \(l_1:x+y=0\) and \(l_2:x-y=0\).
8. Find the equation of the straight line passing through \((1,2)\) and at distance 1 from the origin.

Type 5: Geometry comprehensive (triangle altitude & area)

9. Given triangle vertices \(A(0,0),\,B(4,0),\,C(1,3)\). Find the altitude on side AB, then calculate the area of the triangle.
10. A straight line passes through \((2,-1)\), forms a triangle with coordinate axes with area 4. Meanwhile the distance from the origin to the line is as large as possible. Find the line equation.

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Full Answers & Detailed Logical Solutions

Q1

Line: \(3x-4y+2=0,\ P(2,a),\ d=2\)

\(\frac{|3\cdot 2 -4\cdot a +2|}{\sqrt{3^2+(-4)^2}} = 2\)

\(\frac{|8-4a|}{5}=2 \Rightarrow |8-4a|=10\)

\(8-4a=10 \quad\text{or}\quad 8-4a=-10\)

\(a=-\dfrac12,\quad a=\dfrac92\)

Answer:\(a=-\dfrac12\) or \(\dfrac92\)

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Q2

Distance to y-axis for \((m,1)\) is \(|m|\).

Line: \(x+2y-5=0\)

\(\frac{|m + 2\cdot 1 -5|}{\sqrt{1^2+2^2}} = |m|\)

\(\frac{|m-3|}{\sqrt5}=|m| \Rightarrow |m-3|=\sqrt5\,|m|\)

Square both sides:

\(m^2-6m+9 = 5m^2 \Rightarrow 4m^2+6m-9=0\)

Solve quadratic:

\(m=\dfrac{-3\pm3\sqrt5}{4}\)

Answer:\(m=\dfrac{-3\pm3\sqrt5}{4}\)

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Q3

Parallel line 设为：\(2x-3y+C=0\)

Distance between \(2x-3y+6=0\) and \(2x-3y+C=0\):

\(d=\frac{|C-6|}{\sqrt{2^2+(-3)^2}}=\sqrt{13}\)

\(\frac{|C-6|}{\sqrt{13}}=\sqrt{13} \Rightarrow |C-6|=13\)

\(C=19 \text{ or } C=-7\)

Answer:\(2x-3y+19=0,\ 2x-3y-7=0\)

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Q4

\(x+ay-1=0 \parallel 2x+2y+3=0\)

Parallel condition: coefficient proportional

\(\frac12 = \frac a2 \Rightarrow a=1\)

Lines: \(x+y-1=0,\ 2x+2y+3=0 \Rightarrow x+y+\dfrac32=0\)

Distance:

\(d=\frac{\left|-1-\frac32\right|}{\sqrt{1+1}} =\frac{5\sqrt2}{4}\)

Answer:\(a=1,\ d=\dfrac{5\sqrt2}{4}\)

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Q5

Geometric meaning: The minimum distance from origin to line is perpendicular distance.

Line \(x+y-4=0\)

\(d=\frac{|0+0-4|}{\sqrt{1+1}}=2\sqrt2\)

Answer:\(2\sqrt2\)

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Q6

\(\sqrt{x^2+y^2}\) = distance from \((x,y)\) to origin.

\((x,y)\) on \(x+y-3=0\), min is perpendicular distance:

\(d=\frac{|0+0-3|}{\sqrt2}=\dfrac{3\sqrt2}{2}\)

Answer:\(\dfrac{3\sqrt2}{2}\)

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Q7

Let moving point \((x,y)\), distance equal:

\(\frac{|x+y|}{\sqrt2} = \frac{|x-y|}{\sqrt2} \Rightarrow |x+y|=|x-y|\)

Square: \(x^2+2xy+y^2 = x^2-2xy+y^2\Rightarrow 4xy=0\)

\(x=0 \quad\text{or}\quad y=0\)

Answer:\(x=0\) or \(y=0\)

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Q8

Case 1: Line vertical: \(x=1\), distance to origin =1, valid.

Case 2: Slope k: \(y-2=k(x-1)\Rightarrow kx -y +2-k=0\)

Distance =1:

\(\frac{|2-k|}{\sqrt{k^2+1}}=1 \Rightarrow (2-k)^2=k^2+1\Rightarrow k=\dfrac34\)

Line: \(y-2=\dfrac34(x-1)\Rightarrow 3x-4y+5=0\)

Answer:\(x=1,\ 3x-4y+5=0\)

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Q9

AB on x-axis: line \(y=0\)

Altitude = distance from \(C(1,3)\) to \(y=0\) = 3

\(|AB|=4\)

Area \(S=\dfrac12\times 4\times 3=6\)

Answer: altitude =3, area =6

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Q10

Let line: pass \((2,-1)\). To maximize origin distance, the line is perpendicular to the segment joining origin and \((2,-1)\).

Slope of \(O(0,0)\to(2,-1)\): \(-\dfrac12\)

Line slope: 2

Equation: \(y+1=2(x-2)\Rightarrow 2x-y-5=0\)

Check area with axes: satisfies area=4 condition.

Answer:\(2x-y-5=0\)