Chapter 6: Coordinate Geometry of Straight Lines (Set-5)
A line passes through the intersection of 2x+y−5=0 and x−2y+4=0, and also passes through (1,0). The required line is
A 13x−y−13=0
B 13x+y−13=0
C x−13y−13=0
D 13x−y+13=0
Use family of lines: L1+λL2=0. Put (1,0) to find λ=3/5. Simplify the resulting equation to get 13x−y−13=0, which passes through both required conditions.
A line is parallel to 3x−4y+7=0 and its distance from origin is 5, with positive constant term. The line is
A 3x+4y+25=0
B 4x−3y+25=0
C 3x−4y+25=0
D 3x−4y−25=0
Parallel lines keep the same x,y coefficients, so form is 3x−4y+c=0. Distance from origin is |c|/√(3²+4²)=|c|/5=5, so |c|=25. With c>0, c=25.
A line passes through (2,3) and has a normal vector (3,4). Its equation is
A 4x+3y−18=0
B 3x−4y−18=0
C 3x+4y+18=0
D 3x+4y−18=0
Normal vector (3,4) means line is 3(x−2)+4(y−3)=0. Expanding gives 3x−6+4y−12=0, so 3x+4y−18=0. It satisfies the given point.
The foot of perpendicular from P(2,3) to the line x+y−1=0 is
A (1,0)
B (0,1)
C (1,1)
D (0,0)
For ax+by+c=0, foot from (x0,y0) is (x0−a d, y0−b d), where d=(ax0+by0+c)/(a²+b²). Here d=(2+3−1)/2=2, giving foot (0,1).
A point divides A(1,2) and B(4,5) externally in ratio 2:1. The point is
A (3,4)
B (−1,0)
C (7,8)
D (5,6)
External division formula: P((m x2−n x1)/(m−n), (m y2−n y1)/(m−n)). With m=2,n=1 gives x=(8−1)/1=7 and y=(10−2)/1=8.
For lines x−y=0 and x+y−4=0, the angle bisector parallel to x-axis is
A x=2
B y=−2
C x=−2
D y=2
Since both lines have equal normal lengths, bisectors satisfy L1=±L2. Solving gives y=2 and x=2. The one parallel to x-axis is y=2.
For lines 2x−3y+1=0 and 4x+6y−5=0, the value of tanθ between them is
A 5/12
B 3/5
C 12/5
D 5/3
Slopes are m1=−2/(−3)=2/3 and m2=−4/6=−2/3. Then tanθ=|(m1−m2)/(1+m1m2)|=|(4/3)/(1−4/9)|=(4/3)/(5/9)=12/5.
For the lines kx+3y−7=0 and 3x−ky+2=0, the lines are perpendicular for
A k=0 only
B Any real k
C k=±3 only
D k≠0 only
Slopes are m1=−k/3 and m2=−3/(−k)=3/k (when k≠0). Their product is (−k/3)(3/k)=−1. For k=0, one is horizontal and the other vertical, still perpendicular.
A line makes 30° with x-axis, is at distance 3 from origin, and has positive y-intercept. The equation is
A x+√3y+6=0
B √3x−y+6=0
C x−√3y−6=0
D x−√3y+6=0
Slope tan30=1/√3, so line is y=(1/√3)x+c. Distance from origin equals |c|/√(1+m²)=|c|/(2/√3)=3, giving c=2√3. Converting gives x−√3y+6=0.
Lines 3x+4y+5=0 and 3x+4y+k=0 are 2 units apart, with k>5. Then k is
A −5
B 5
C 15
D 25
Distance between parallel lines is |k−5|/√(3²+4²)=|k−5|/5=2. So |k−5|=10 gives k=15 or −5. With k>5, choose 15.
The perpendicular bisector of segment joining (−1,2) and (3,6) is
A x+y−5=0
B x−y−5=0
C x+y+5=0
D x−y+5=0
Midpoint is (1,4). Segment slope is (6−2)/(3−(−1))=1, so perpendicular slope is −1. Through (1,4): y−4=−(x−1) gives x+y−5=0.
A line is given by x=2+3t, y=−1+2t. Its y-intercept is
A 7/3
B −3/7
C 3/7
D −7/3
At y-intercept, x=0. So 0=2+3t gives t=−2/3. Then y=−1+2(−2/3)=−1−4/3=−7/3. Hence intercept point is (0,−7/3).
The symmetric line (x+1)/2 = (y−3)/−1 in general form is
A 2x+y−5=0
B x−2y−5=0
C x+2y−5=0
D x+2y+5=0
Put (x+1)/2=(y−3)/−1=t. Then x=−1+2t, y=3−t. Eliminate t: t=(x+1)/2, so y=3−(x+1)/2=(5−x)/2, giving x+2y−5=0.
Points (1,k), (3,4), (5,8) are collinear. The value of k is
A 0
B 2
C 4
D −2
Slope between (3,4) and (5,8) is 4/2=2. For collinearity, slope between (1,k) and (3,4) must be 2: (4−k)/2=2, so 4−k=4 and k=0.
The line x+2y−6=0 divides segment joining A(0,0) and B(6,3) in ratio
A 2:1
B 1:2
C 1:1
D 3:1
A point on AB is (6t,3t). Substitute in line: 6t+2(3t)−6=0 →12t=6 →t=1/2. That is the midpoint, so ratio is 1:1.
The area of triangle formed by line x+y=6 with coordinate axes is
A 12
B 18
C 36
D 9
x-intercept is 6 and y-intercept is 6. Triangle area with axes is (1/2)×6×6=18. This uses the intercepts as base and height.
A line x/a + y/b = 1 forms a triangle of area 18 with axes, and x-intercept a=6. Then b equals
A 3
B 9
C 12
D 6
Area formed with axes is (1/2)|ab|. Given area 18 and a=6: (1/2)(6b)=18 →3b=18 →b=6. So y-intercept is 6.
A line passes through (2,1) and has equal intercepts on both axes. The line is
A x−y−3=0
B x+y+3=0
C x+y−3=0
D x−y+3=0
Equal intercepts mean x+y=a. Passing through (2,1) gives 2+1=a=3. So equation is x+y=3, or x+y−3=0.
A line parallel to y=x lies above it and is at distance 3/√2 from origin. The equation is
A y=x−3
B y=x+3
C y=−x+3
D y=−x−3
Lines parallel to y=x have form y=x+c. Distance from origin is |c|/√2. Given 3/√2 gives |c|=3. “Above” means c>0, so c=3 and y=x+3.
A line passes through (4,1) and makes 45° with positive x-axis. The equation is
A y=−x+5
B x=4
C y=1
D y=x−3
Inclination 45° means slope m=tan45=1. Through (4,1): y−1=1(x−4) gives y=x−3. Other options either wrong slope or wrong passing condition.
A line passes through intersection of x+y−4=0 and x−y−2=0 and is parallel to 2x+y=0. The line is
A 2x+y−7=0
B 2x+y+7=0
C x+2y−7=0
D 2x−y−7=0
Intersection of x+y=4 and x−y=2 gives (3,1). A line parallel to 2x+y=0 has same coefficients: 2x+y+c=0. Put (3,1): 6+1+c=0 → c=−7.
For 3x+4y+k=0 passing through (2,−1), k equals
A 2
B −10
C −2
D 10
Substitute (2,−1): 3(2)+4(−1)+k=0 →6−4+k=0 →k=−2. This ensures the point lies exactly on the line.
The perpendicular from P(2,3) to line 3x−4y+7=0 has equation
A 4x−3y−17=0
B 3x+4y−17=0
C 4x+3y+17=0
D 4x+3y−17=0
Given line slope is 3/4, so perpendicular slope is −4/3. Through (2,3): y−3=(−4/3)(x−2). Rearranging gives 4x+3y−17=0.
The foot of perpendicular from P(2,3) to 3x−4y+7=0 is
A (79/25,47/25)
B (47/25,79/25)
C (2,3)
D (0,0)
Use d=(ax0+by0+c)/(a²+b²) with a=3,b=−4,c=7. d=(6−12+7)/25=1/25. Foot is (2−3/25, 3−(−4)(1/25))=(47/25,79/25).
The foot of perpendicular from origin to line x+2y−6=0 is
A (12/5,6/5)
B (6,12)
C (6/5,12/5)
D (0,0)
For origin (0,0), d=c/(a²+b²)=−6/5. Foot is (−a d, −b d) = (6/5, 12/5). This point gives the shortest distance from origin.
The line midway between 2x+3y−6=0 and 2x+3y+12=0 is
A 2x+3y+3=0
B 2x+3y−3=0
C 2x−3y+3=0
D 3x+2y+3=0
For parallel lines ax+by+c1=0 and ax+by+c2=0, the midline is ax+by+(c1+c2)/2=0. Here (−6+12)/2=3, so 2x+3y+3=0.
Lines x+ky−3=0 and 2x+2ky+1=0 are parallel. If their distance is 1 and k>0, k is
A √5/2
B 3√5
C 5/2
D 3√5/2
Divide second by 2: x+ky+1/2=0. Distance is |(−3)−(1/2)|/√(1+k²)=(7/2)/√(1+k²)=1. So √(1+k²)=7/2 → k²=45/4 → k=3√5/2 (k>0).
A line has intercepts in ratio 2:3 and passes through (2,1). The equation is
A 2x+3y−8=0
B 3x−2y−8=0
C 3x+2y−8=0
D 3x+2y+8=0
Let intercepts be a=2t, b=3t. Line: x/(2t)+y/(3t)=1 → 3x+2y=6t. Using (2,1): 6+2=8=6t gives t=4/3, so 3x+2y=8.
A line through (1,2) makes 45° with x−y=0 and is not vertical. The line is
A x=1
B y=2
C y=x+1
D x+y=3
Line x−y=0 has slope 1 (45°). A line making 45° with it has inclination 0° or 90°, so slope 0 or undefined. Not vertical means choose slope 0: y=2 through (1,2).
A line passes through (1,2) and has x-intercept and y-intercept sum 6, with y-intercept greater. The equation is
A x+y−3=0
B x+2y−4=0
C 2x+y−6=0
D 2x+y−4=0
Let intercepts be a and b with a+b=6 and b>a. Passing through (1,2): 1/a + 2/b =1. Solutions give (a,b)=(2,4) or (3,3). With b>a, choose (2,4): x/2 + y/4=1 → 2x+y−4=0.
Lines 2x+3y−6=0, x−y+1=0, and kx+2y−3=0 are concurrent. Then k is
A 1/3
B −3
C −1/3
D 3
Solve first two: from x−y=−1 gives x=y−1. Substitute in 2x+3y=6: 2(y−1)+3y=6 →5y=8 →y=8/5, x=3/5. Put in third: k(3/5)+16/5−3=0 → (3k+1)/5=0 → k=−1/3.
Lines x+y−1=0, 2x+3y−4=0, and 3x+ky−5=0 are concurrent. Then k is
A 4
B 2
C −4
D 1
Intersection of first two: y=1−x. Substitute in 2x+3(1−x)=4 →2x+3−3x=4 →x=−1, y=2. Put into third: 3(−1)+k(2)−5=0 →2k−8=0 →k=4.
The pair of lines 5x²+2hxy−5y²=0 represents perpendicular lines for
A h=0 only
B h=±5 only
C No real h
D All real h
For ax²+2hxy+by²=0, product of slopes is a/b. Perpendicular means product −1, so a/b=−1 → a+b=0. Here 5+(−5)=0, so always perpendicular; discriminant is always positive.
For x²−5xy+6y²=0, the product of slopes of the two lines is
A 6
B −1/6
C 1/6
D −6
Put y=mx in ax²+2hxy+by²=0 gives b m²+2h m+a=0. Product of roots m1m2=a/b. Here a=1, b=6, so m1m2=1/6.
For x²−5xy+6y²=0, the sum of slopes of the two lines is
A 5/6
B 6/5
C −5/6
D −6/5
From b m²+2h m+a=0, sum of roots m1+m2=−2h/b. Here 2h=−5 so h=−5/2. Thus sum = −2(−5/2)/6 = 5/6.
For 2x²+2hxy+2y²=0 to represent two real distinct lines, the smallest positive integer h is
A 2
B 1
C 3
D 4
Condition for real distinct pair is h²>ab where equation is ax²+2hxy+by²=0. Here a=2, b=2 so ab=4. Need h²>4, so smallest positive integer is 3.
A line has equal intercepts and passes through (1,3). The line is
A x+y−3=0
B x−y−4=0
C x−y+4=0
D x+y−4=0
Equal intercepts mean x+y=a. Passing through (1,3) gives 1+3=a=4. So equation is x+y=4, i.e., x+y−4=0.
The locus of points whose distances from y=1 and y=7 have sum 10 is
A y=5 only
B y=9 or y=−1
C y=1 or y=7
D x=9 or x=−1
Sum |y−1|+|y−7| equals 6 for 1≤y≤7, not 10. For y≥7, sum=2y−8=10 gives y=9. For y≤1, sum=8−2y=10 gives y=−1. So two lines.
A line is at distance 2 from 3x−4y+1=0 on the same side as origin. The line is
A 3x−4y−9=0
B 4x−3y+11=0
C 3x−4y+11=0
D 3x+4y+11=0
Parallel lines: 3x−4y+c=0. Distance is |c−1|/5=2, so |c−1|=10 giving c=11 or −9. Check origin sign: original gives +1; choose c=11 to keep same side.
The signed distance of origin from x+2y−6=0 is
A 6/√5
B −√5/6
C 0
D −6/√5
Signed distance from (x0,y0) to ax+by+c=0 is (ax0+by0+c)/√(a²+b²). For origin, it is (−6)/√(1+4)=−6/√5. Negative shows origin lies opposite normal direction.
Points equidistant from lines x−2y=0 and 2x+y=0 lie on
A x+3y=0 or 3x−y=0
B x−3y=0 or 3x+y=0
C x+y=0 or x−y=0
D x=0 or y=0
Both lines have equal normal length √5, so angle bisectors satisfy L1=±L2. Solve x−2y = 2x+y giving x+3y=0, and x−2y = −(2x+y) giving 3x−y=0.
The width of the strip between 2x−3y+4=0 and 4x−6y−8=0 is
A 4/√13
B 8/√13
C 8/13
D √13/8
Make coefficients same: divide second by 2 → 2x−3y−4=0. Distance between parallel lines is |4−(−4)|/√(2²+(−3)²)=8/√13. This is the constant strip width.
A line passes through (2,3) and has intercepts equal in magnitude but opposite signs. The equation is
A x+y−1=0
B x−y−1=0
C x+y+1=0
D x−y+1=0
Opposite equal intercepts imply a=−b. Then x/a + y/(−a)=1 gives x−y=a. Using (2,3): 2−3=−1, so a=−1, hence x−y+1=0.
A line through (1,2) is perpendicular to x+2y−5=0 and also passes through (3,k). The value of k is
A 4
B 2
C 6
D 0
Slope of x+2y−5=0 is −1/2, so perpendicular slope is 2. Slope between (1,2) and (3,k) is (k−2)/2. Set (k−2)/2=2 → k−2=4 → k=6.
For line 3x−4y+c=0, the distance from point (2,1) is 1 and c<0. Then c is
A 3
B −7
C −2
D 7
Distance is |3(2)−4(1)+c|/5 = |2+c|/5 = 1. So |2+c|=5 giving c=3 or c=−7. With c<0, choose −7.
A line passes through intersection of 2x−y=0 and x+2y−6=0, and has x-intercept 4. The line is
A 6x−7y−24=0
B 7x+6y−24=0
C 6x+7y−24=0
D 6x+7y+24=0
Intersection: y=2x, substitute in x+4x−6=0 gives x=6/5, y=12/5. Line through (4,0) and (6/5,12/5) has slope −6/7. Equation becomes 7y=−6x+24 → 6x+7y−24=0.
A line with slope 2 passes through (1,3). Its general form with integer coefficients is
A 2x+y+1=0
B x−2y+1=0
C y−2x+1=0
D 2x−y+1=0
Slope 2 means y=2x+c. Use (1,3): 3=2(1)+c gives c=1. So y=2x+1 → 2x−y+1=0. It matches the point and slope.
A point P lies on line 4x+3y−12=0 and has x-coordinate 3. Then y-coordinate of P is
A 0
B 1
C 4
D −4
Substitute x=3 in 4x+3y−12=0: 12+3y−12=0 gives 3y=0 so y=0. So the point is (3,0) on the line.
The distance between point (0,0) and line 5x+12y−13=0 is
A 13
B 13/5
C 1
D 5/13
Distance from origin is |c|/√(a²+b²). Here |−13|/√(25+144)=13/13=1. So the perpendicular distance from (0,0) to the line is exactly 1 unit.
A line through (2,−1) is parallel to 4x−6y+9=0. The correct equation is
A 4x−6y+14=0
B 6x−4y−14=0
C 4x+6y−14=0
D 4x−6y−14=0
Parallel lines keep coefficients of x and y. So form is 4x−6y+c=0. Substitute (2,−1): 8+6+c=0 gives c=−14. Hence 4x−6y−14=0.