Chapter 6: Coordinate Geometry of Straight Lines (Set-4)
The line through (2,3) and (8,15) has slope
A 3
B 1/2
C −2
D 2
Slope m=15−38−2=126=2m=8−215−3=612=2. Positive slope means the line rises to the right at a rate of 2 units in y per 1 unit in x.
A line with slope −2 passing through (1,4) is
A y+4=−2(x−1)
B y−4=2(x−1)
C y−4=−2(x−1)
D y−1=−2(x−4)
Use point-slope form y−y1=m(x−x1)y−y1=m(x−x1). With point (1,4) and slope −2, it becomes y−4=−2(x−1)y−4=−2(x−1), which is correct and unique.
The equation of line passing through (0,−3) and (6,0) is
A x−2y−6=0
B x+2y+6=0
C 2x−y−6=0
D x−2y+6=0
Slope m=0−(−3)6−0=3/6=1/2m=6−00−(−3)=3/6=1/2. Through (0,−3): y+3=12x⇒x−2y−6=0y+3=21x⇒x−2y−6=0.
Which line is parallel to 3x+4y−8=0?
A 4x+3y+5=0
B 6x+8y+5=0
C 3x−4y+5=0
D 8x+6y+5=0
Parallel lines have proportional x and y coefficients. (3,4) and (6,8) are proportional, so option A is parallel. Different constant ensures it is not the same line.
Which line is perpendicular to 2x−5y+1=0?
A 2x−5y−3=0
B x+2y+3=0
C 5x+2y+3=0
D 5x−2y+3=0
Slope of given line is −A/B=−2/(−5)=2/5−A/B=−2/(−5)=2/5. Perpendicular slope is −5/2−5/2. Line 5x+2y+3=0 has slope −5/2−5/2, so it’s perpendicular.
A line cuts x-axis at (−4,0) and y-axis at (0,2). Its equation is
A x/−4 + y/2 = 1
B x/4 + y/2 = 1
C x/−4 − y/2 = 1
D x/2 + y/4 = 1
Intercept form is x/a+y/b=1x/a+y/b=1. Here a=−4 and b=2. So equation becomes x/(−4)+y/2=1x/(−4)+y/2=1, matching option A.
The y-intercept of 5x−2y+10=0 is
A −5
B 2
C 5
D −2
Put x=0: −2y+10=0 ⇒ y=5? Wait carefully: −2y = −10 ⇒ y = 5. But y-intercept is 5, not −5. So check again: 5x−2y+10=0 gives y= (5/2)x+5. Hence intercept is 5.
For 5x−2y+10=0, slope is
A 5/2
B −5/2
C 2/5
D −2/5
In Ax+By+C=0Ax+By+C=0, slope m=−A/Bm=−A/B. Here A=5, B=−2, so m=−5/(−2)=5/2m=−5/(−2)=5/2. The line rises steeply to the right.
The acute angle between lines with slopes 1 and 3 is given by tanθ =
A 1/5
B 1/2
C 2
D 5
tanθ=∣m1−m21+m1m2∣=∣1−31+3∣=24=1/2tanθ=1+m1m2m1−m2=1+31−3=42=1/2. So correct tanθ is 1/2, not 1/5.
If lines have slopes 2 and 1/2, then tanθ equals
A 1
B 0
C 3/4
D 5/3
tanθ=∣2−1/21+2⋅1/2∣=∣3/22∣=3/4tanθ=1+2⋅1/22−1/2=23/2=3/4. So tanθ is 3/4, not 3/5.
The distance from (3,4) to line x−2y+1=0 is
A 4/√5
B 6/√5
C 2/√5
D √5/2
Distance =∣x1−2y1+1∣12+(−2)2=∣3−8+1∣5=45=12+(−2)2∣x1−2y1+1∣=5∣3−8+1∣=54. So correct is 4/√5.
Distance between 4x+3y−9=0 and 4x+3y+6=0 is
A 3/5
B 15/√25
C 15/5
D 15/5
Distance = |c1−c2|/√(a²+b²) = |−9−6|/√(16+9)=15/5=3. So 15/5 is the simplest correct form.
The intersection of x+y=6 and x−y=2 is
A (2,4)
B (3,3)
C (4,2)
D (6,2)
Add equations: 2x=8 so x=4. Substitute into x+y=6 gives y=2. So intersection point is (4,2), where both lines meet.
Point (1,2) lies on which line?
A 2x−y=0
B x+2y=5
C x−2y=0
D 3x+y=10
Substitute (1,2). For B: 1+4=5, satisfied. For A: 2−2=0 also satisfied; wait A also works. So this question has two correct options.
A line through (−1,3) parallel to y=2x+1 is
A y=2x+5
B y=2x+1
C y=−2x+5
D y=−2x+1
Parallel lines have same slope 2. Use y=2x+c through (−1,3): 3=2(−1)+c ⇒ c=5. So equation is y=2x+5.
A line through (4,−2) perpendicular to y=−x+3 is
A y=−x−6
B y=x+6
C y=x−6
D y=−x+6
Given line slope is −1. Perpendicular slope is 1. Through (4,−2): y+2 = 1(x−4) ⇒ y = x−6.
The midpoint of (−3,5) and (7,−1) is
A (2,2)
B (1,2)
C (2,1)
D (4,2)
Midpoint is (−3+72,5+(−1)2)=(2,2)(2−3+7,25+(−1))=(2,2). It lies halfway in both x and y directions.
If P divides A(0,0) and B(6,4) in ratio 1:2, then P is
A (4,8/3)
B (1,1)
C (2,4/3)
D (3,2)
Section formula internal: P=(1⋅6+2⋅03,1⋅4+2⋅03)=(2,4/3)P=(31⋅6+2⋅0,31⋅4+2⋅0)=(2,4/3). Ratio 1:2 means closer to A.
Area of triangle with (0,0), (4,0), (0,3) is
A 12
B 6
C 3
D 24
This is a right triangle with base 4 and height 3. Area = 1/2 × 4 × 3 = 6 square units, matching option A.
Centroid of (1,1), (4,1), (1,7) is
A (3,4)
B (2,4)
C (3,3)
D (2,3)
Centroid is average of coordinates: (1+4+13,1+1+73)=(2,3)(31+4+1,31+1+7)=(2,3). It is the common intersection point of medians.
A line in symmetric form (x−2)/3=(y+1)/−2 passes through
A (3,−2)
B (2,1)
C (2,−1)
D (−2,1)
Symmetric form shows the line passes through (x1,y1) = (2,−1). Direction ratios are (3,−2). So point (2,−1) must lie on the line.
For parametric x=−1+t, y=2+3t, direction ratios are
A (1,3)
B (3,1)
C (−1,3)
D (1,−3)
In parametric form, coefficients of t give direction ratios. Here x changes by 1 per t and y changes by 3 per t, so direction ratios are (1,3).
Line through intersection of x−y=0 and x+y−4=0 is
A (x−y)/(x+y−4)=0
B (x−y)+λ(x+y−4)=0
C (x−y)−(x+y−4)=0
D (x−y)λ(x+y−4)=0
Any line passing through intersection of L1=0 and L2=0 is given by L1 + λL2 = 0. This ensures the common intersection point satisfies the family.
For line y=mx+c, if c=0, then the line
A passes origin
B vertical line
C horizontal line
D parallel y-axis
If c=0, then y=mx. Substituting x=0 gives y=0, so the line always passes through the origin regardless of the slope m.
The slope of line joining (a,b) and (a,−b) is
A 0
B −1
C Undefined
D 1
Both points have same x=a, so the line is vertical. Vertical line means Δx=0, hence slope Δy/Δx is undefined.
The slope of line joining (−a,b) and (a,b) is
A 1
B −1
C Undefined
D 0
Both points share same y=b, so the line is horizontal. Horizontal lines have Δy=0, so slope is 0.
Distance from (1,−1) to line 3x+4y=0 is
A 1/5
B 1
C 7/5
D 5/7
Distance =∣3(1)+4(−1)+0∣9+16=∣−1∣5=1/5=9+16∣3(1)+4(−1)+0∣=5∣−1∣=1/5. This is the perpendicular distance from the point to the line.
The line 2x+3y+6=0 shifted up by 2 units becomes
A 2x+3y=6
B 2x+3y+12=0
C 2x+3y=0
D 2x+3y=−6
Shifting a line “up” changes its constant term in a way that depends on its orientation; it’s not simply adding 6. This needs careful transformation, so option B is not reliable.
A point is 3 units from line y=1. Its locus is
A y=3 only
B y=4 or y=−2
C x=4 or x=−2
D x=3 only
Points at distance 3 from horizontal line y=1 lie on two parallel lines: y = 1+3 = 4 and y = 1−3 = −2.
Distance between lines x=2 and x=−5 is
A 3
B 10
C 7
D 5
These are vertical parallel lines. Their distance is horizontal difference |2−(−5)| = 7 units, which is the shortest distance between them.
For line x cosα + y sinα = p, p must be
A nonnegative
B negative always
C zero always
D imaginary
In normal form, p is the perpendicular distance from origin to the line, taken as nonnegative. Any sign change is adjusted by changing the angle α direction.
If a line has normal vector (0,1), it is
A y=x line
B horizontal
C y=−x line
D vertical
Normal (0,1) points along y-axis, so the line is perpendicular to y-axis, meaning it is horizontal. Such lines have equation y = constant.
If direction vector is (0,1), the line is
A slope 1
B horizontal
C vertical
D slope −1
Direction vector (0,1) means x does not change while y changes, so the line goes straight up/down. That is a vertical line with undefined slope.
The line y=2x+1 makes angle θ with x-axis where tanθ is
A 1/2
B −2
C 0
D 2
For a line, slope m = tanθ where θ is inclination with positive x-axis. Here slope is 2, so tanθ = 2, giving an acute angle less than 90°.
If slopes m1 and m2 satisfy 1+m1m2=0, lines are
A perpendicular
B parallel
C coincident
D no relation
Condition 1 + m1m2 = 0 gives m1m2 = −1, which is exactly the perpendicular condition for two non-vertical lines.
For lines Ax+By+C=0 and Bx−Ay+K=0, they are
A parallel
B same line
C perpendicular
D horizontal only
First line slope is −A/B. Second line slope is −B/(−A)=B/A. Product is (−A/B)·(B/A)=−1, so the lines are perpendicular.
If point (2,3) is reflected in x-axis, image is
A (2,−3)
B (−2,3)
C (−2,−3)
D (3,2)
Reflection in x-axis keeps x same and changes sign of y. So (2,3) becomes (2,−3). Distance to x-axis remains same.
If point (−4,1) is reflected in y-axis, image is
A (−1,4)
B (4,−1)
C (−4,−1)
D (4,1)
Reflection in y-axis changes sign of x while y remains unchanged. So (−4,1) becomes (4,1). The y-coordinate stays the same.
A point equidistant from lines y=x and y=−x lies on
A x=2y line
B x-axis or y-axis
C y=3 line
D y=2x line
Angle bisectors of y=x and y=−x are the coordinate axes. Points equidistant from the two lines lie on these bisectors, i.e., x=0 or y=0.
The area of triangle formed by (0,0), (2,2), (4,4) is
A 0
B 4
C 8
D 2
The points lie on the line y=x and are collinear. Collinear points do not form a triangle with height, so area is zero.
Equation of line through origin and perpendicular to x+3y=0 is
A x−3y=0
B x+3y=0
C 3x−y=0
D y=3x
Given line slope is −1/3. Perpendicular slope is 3. Through origin gives y=3x, or 3x−y=0, which is the required line.
If a line is at distance 2 from origin and parallel to y=0, it is
A y=2 or y=−2
B x=2 or x=−2
C y=2 only
D x=2 only
Parallel to y=0 means horizontal lines. Distance 2 from origin gives two such lines above and below: y=2 and y=−2.
A line passes through (1,2) and is parallel to x-axis. It is
A x=1
B y=2
C y=x+1
D x+y=3
Parallel to x-axis means horizontal line, so y is constant. Passing through (1,2) fixes y=2. This is independent of x.
A line passes through (1,2) and is parallel to y-axis. It is
A y=1
B x=2
C x=1
D y=2
Parallel to y-axis means vertical line, so x is constant. Passing through (1,2) fixes x=1. y can be any value on the line.
The equation of line joining (0,5) and (5,0) is
A x+y−5=0
B x−y−5=0
C x+y+5=0
D x−y+5=0
Slope is (0−5)/(5−0)=−1. Using intercepts 5 and 5 gives intercept form x/5 + y/5 =1 ⇒ x+y=5 ⇒ x+y−5=0.
The shortest distance from point (a,b) to x-axis is
A √(a²+b²)
B |a|
C |a+b|
D |b|
x-axis is y=0. Perpendicular distance from (a,b) to y=0 is |b−0| = |b|. Only y-coordinate matters for vertical distance.
The shortest distance from point (a,b) to y-axis is
A √(a²+b²)
B |a|
C |b|
D |a−b|
y-axis is x=0. Perpendicular distance from (a,b) to x=0 is |a−0| = |a|. Only x-coordinate matters for horizontal distance.
If line ax+by+c=0 passes through origin, then
A c=0
B a=0
C b=0
D a=b
Substituting (0,0) gives a·0 + b·0 + c = 0 ⇒ c=0. So any line through origin in general form must have constant term zero.
If two lines are perpendicular, their direction vectors have
A same magnitude
B same direction
C zero dot product
D equal components
Two directions are perpendicular when their dot product is zero. This matches the idea that cos90°=0, so the scalar product becomes zero.
A line passes through (0,0) and has direction ratios (3,4). Its slope is
A 3/4
B −3/4
C −4/3
D 4/3
Direction ratios (a,b) give slope m=b/a. Here m=4/3. Through origin means equation is y=(4/3)x, showing rise 4 for run 3.