Mixed problems on lines

Harder questions are several small steps: turn each condition into an equation, then solve them together.

Concurrent lines (three or more through one point). Find where two of them meet, then make the third pass through that point. x + y = 4 and x – y = 2 meet at (3, 1); for 2x + ky = 10 to pass through it, 6 + k = 10, so k = 4.

Image (mirror reflection) of a point in a line. The line is the perpendicular bisector of the point and its image. Image of P(4, 1) in x – y + 1 = 0:

• Two conditions: the mid-point of P and its image (α, β) lies on the line, and the slope from P to the image times the line's slope is –1.
• Shortcut: put P into the line: 4 – 1 + 1 = 4; with A² + B² = 2, move P by –2 × 4/2 = –4 times (A, B) = (1, –1).
• Image (0, 5). Check: the mid-point (2, 3) satisfies 2 – 3 + 1 = 0.

A ray of light reflected by the x-axis: the ray from (1, 3) bounces off the x-axis and passes through (5, 1). Reflect (5, 1) to (5, –1); the incoming ray points straight at it. The line from (1, 3) to (5, –1) has slope –1 and meets the x-axis at (4, 0), the point of reflection.

In what ratio does a line divide a segment? Take the ratio as k : 1, write the dividing point with the section formula, and put it into the line. For 2x + y = 7 and the segment from (1, 1) to (5, 3):

• The point is ((5k + 1)/(k + 1), (3k + 1)/(k + 1)).
• Putting it in: 2(5k + 1) + (3k + 1) = 7(k + 1), so 6k = 4 and k = 2/3.
• The ratio is 2 : 3. Check: the point (13/5, 9/5) gives 26/5 + 9/5 = 7.

Lines at a given angle to a line. Lines through (2, 1) making 45° with 3x – y = 0 (slope 3): |(m – 3)/(1 + 3m)| = 1 gives two cases.

• m – 3 = 1 + 3m gives m = –2: 2x + y – 5 = 0.
• m – 3 = –(1 + 3m) gives m = 1/2: x – 2y = 0.

Equal distances from two lines. Points equally far from 3x – 4y + 1 = 0 and 4x + 3y – 2 = 0 (both have √(A² + B²) = 5) satisfy 3x – 4y + 1 = ±(4x + 3y – 2). That gives x + 7y – 3 = 0 and 7x – y – 1 = 0: the two lines that bisect the angles between them. They are straight lines, and they are perpendicular to each other.