Chapter 6: Coordinate Geometry of Straight Lines (Set-2)
The slope of line y = 7 is
A 1
B −1
C 0
D Undefined
In y = 7, y is constant for all x, so Δy = 0. Hence slope m = Δy/Δx = 0, meaning the line is horizontal.
The slope of line x = −9 is
A Undefined
B 0
C 1
D −1
In x = −9, x stays constant. Slope is Δy/Δx, but Δx = 0 for any two points on this line, so slope is undefined.
A line with slope −3 is
A Rising right
B Falling right
C Horizontal
D Vertical
Negative slope means as x increases, y decreases. So the line goes downward from left to right, commonly described as “falling right.”
The line y = 2x + 5 cuts y-axis at
A 2
B −2
C 5
D −5
The y-intercept occurs at x = 0. Substituting gives y = 2(0) + 5 = 5, so the line crosses the y-axis at 5.
The line y = −4x + 1 has slope
A 4
B 1
C −1
D −4
In y = mx + c, slope equals the coefficient of x. Here the coefficient is −4, so the line slopes downward as x increases.
A line through origin must satisfy
A c = 0
B m = 0
C A = 0
D B = 0
For y = mx + c to pass through (0,0), substitute x=0, y=0 giving 0 = c. Therefore, c must be 0.
The x-intercept of y = 3x − 6 is
A −2
B 6
C 2
D −6
x-intercept occurs when y = 0. Put 0 = 3x − 6, so 3x = 6 and x = 2, giving intercept at (2,0).
In intercept form x/a + y/b = 1, “a” is
A y-intercept
B x-intercept
C slope value
D distance value
Put y = 0 in x/a + y/b = 1 gives x/a = 1 so x = a. Thus a is the x-intercept of the line.
In intercept form x/a + y/b = 1, “b” is
A y-intercept
B x-intercept
C slope inverse
D angle value
Put x = 0 in x/a + y/b = 1 gives y/b = 1 so y = b. Hence b is the y-intercept.
Equation through (4,−1) with slope 2 is
A y−1=2(x−4)
B y+1=−2(x−4)
C y+1=2(x−4)
D y−1=−2(x−4)
Point-slope form is y − y1 = m(x − x1). With (x1,y1)=(4,−1) and m=2, we get y + 1 = 2(x − 4).
Slope of line through (1,5) and (4,5) is
A 1
B −1
C Undefined
D 0
Both points have same y-value, so the segment is horizontal. Slope m = (5−5)/(4−1) = 0/3 = 0.
Slope of line through (2,1) and (2,7) is
A 0
B 1
C Undefined
D −1
Both points have same x-value, so the line is vertical. Δx = 0, so slope m = Δy/Δx is undefined.
In Ax+By+C=0, slope equals
A −A/B
B A/B
C −B/A
D B/A
Rewrite Ax + By + C = 0 as y = (−A/B)x − C/B. The coefficient of x is the slope, so m = −A/B.
For 5x − y + 2 = 0, slope is
A −5
B 5
C 1/5
D −1/5
Here A=5 and B=−1. Slope m = −A/B = −5/(−1) = 5. So the line rises steeply to the right.
A line parallel to 2x+3y=7 is
A 3x+2y=1
B 2x−3y=1
C 2x+3y=1
D x+ y=1
Parallel lines have proportional coefficients of x and y. Keeping same 2 and 3 but changing constant gives a distinct parallel line.
A line perpendicular to y = 3x + 1 has slope
A 1/3
B −3
C 3
D −1/3
Given slope m=3. Perpendicular slope m⊥ satisfies m·m⊥ = −1, so m⊥ = −1/3.
Angle between slopes m and −1/m is
A 90°
B 0°
C 30°
D 45°
If slopes are negative reciprocals, product is −1, which means the lines are perpendicular. Perpendicular lines meet at 90°.
If m1 = 1 and m2 = 0, tanθ is
A 0
B −1
C 1
D Undefined
tanθ = |(m1−m2)/(1+m1m2)| = |(1−0)/(1+0)| = 1. So the acute angle θ is 45°.
If tanθ = 0, lines are
A perpendicular
B parallel
C intersecting 60°
D always same
tanθ = 0 means θ = 0° (acute angle). That happens when slopes are equal, so the lines are parallel or coincident.
Acute angle between two lines is always
A < 90°
B > 90°
C = 90°
D = 180°
By definition, the acute angle is the smaller angle between two intersecting lines, and it is always less than 90°.
Normal vector for 7x+2y−5=0 is
A (2,7)
B (−7,2)
C (7,2)
D (7,−2)
For Ax+By+C=0, the normal direction is along (A,B). So here the normal vector is (7,2), perpendicular to the line.
Distance from (1,1) to x+y−2=0 is
A 0
B 1/√2
C √2
D 2
Substitute (1,1): 1+1−2 = 0, so the point lies on the line. Distance from a point on the line to the line is zero.
Distance from origin to x−4=0 is
A 0
B 2
C Undefined
D 4
x−4=0 means x=4, a vertical line. The perpendicular distance from (0,0) to x=4 is |0−4| = 4.
Distance from origin to y+6=0 is
A 0
B 3
C 6
D 12
y+6=0 means y=−6, a horizontal line. Distance from (0,0) to y=−6 is |0−(−6)| = 6.
Distance from (2,0) to y=3 is
A 1
B 3
C 2
D 5
Line y=3 is horizontal. Distance from point (2,0) is vertical difference |0−3| = 3. x-coordinate does not affect distance here.
Distance between 2x+3y+1=0 and 2x+3y−5=0 is
A 6/√13
B 1
C 2
D √13/6
For ax+by+c1=0 and ax+by+c2=0, distance = |c1−c2|/√(a²+b²). Here |1−(−5)|=6 and √(4+9)=√13.
Distance between y=2 and y=9 is
A 9
B 11
C 7
D 18
These are horizontal parallel lines. Distance equals the absolute difference in y-values: |2−9| = 7.
A line passing through (0,5) is sure to have
A x-intercept 5
B y-intercept 5
C slope 5
D normal 5
The point (0,5) lies on y-axis. Any line through it crosses y-axis at y=5, so y-intercept must be 5.
A line passing through (−3,0) is sure to have
A y-intercept −3
B slope −3
C angle −3
D x-intercept −3
The point (−3,0) lies on x-axis. Therefore the line crosses x-axis at x=−3, making x-intercept −3.
A line with equation y = mx has
A vertical line
B zero slope
C passes origin
D no intercept
In y=mx, constant term is 0. Substituting x=0 gives y=0, so the line always passes through the origin.
A line parallel to y-axis has equation
A x = constant
B y = constant
C y = mx
D x + y = 0
Any line parallel to y-axis is vertical and keeps x fixed. Hence its equation is x = k for some constant k.
A line parallel to x-axis has equation
A x = constant
B y = mx + c
C y = constant
D Ax + By = 0
Any line parallel to x-axis is horizontal and keeps y fixed. So its equation is y = k for some constant k.
For points (1,2), (2,4), (3,6), they are
A non-collinear
B collinear
C perpendicular set
D parallel set
Slope between (1,2) and (2,4) is 2, and between (2,4) and (3,6) is also 2. Equal slopes show all three points lie on one line.
In 2D, direction ratios of (4,−2) are
A −4, 2
B 2, −4
C 1, 2
D 4, −2
Direction ratios can be any numbers proportional to a direction vector. If direction vector is (4,−2), then (4,−2) itself is a valid pair.
A direction vector for slope 1 is
A (1,−1)
B (0,1)
C (1,1)
D (1,0)
Slope m = Δy/Δx. For slope 1, we can take Δx=1 and Δy=1, giving direction vector (1,1).
Symmetric form in 2D can be written as
A (x−x1)/a = (y−y1)/b
B y = mx + c
C Ax + By + C = 0
D x/a + y/b = 1
A line through (x1,y1) with direction ratios (a,b) can be expressed as (x−x1)/a = (y−y1)/b, showing equal parameter along direction.
Parametric form uses parameter “t” as
A fixed constant
B running variable
C y-intercept
D slope only
Parametric form expresses x and y as functions of a parameter t, which varies over real numbers to generate all points on the line.
A line through intersection of L1 and L2 is
A L1 − L2 = 0
B L1·L2 = 0
C L1 + λL2 = 0
D L1/L2 = 0
If L1=0 and L2=0 are two lines, their intersection point satisfies both. Any line passing through that point is given by L1 + λL2 = 0.
Pair of lines through origin can be written as
A y = mx + c
B Ax + By + C = 0
C x/a + y/b = 1
D ax²+2hxy+by²=0
A homogeneous second-degree equation in x and y represents a pair of straight lines through the origin, because it remains satisfied under scaling of (x,y).
Reflection of a point about a line keeps
A equal perpendicular distance
B equal x-coordinate
C equal y-intercept
D equal slope
Reflection across a line makes the line the perpendicular bisector of the segment joining the point and its image. Hence both points have equal perpendicular distance from the line.
Locus of points at fixed distance from a line forms
A circle
B parabola
C two parallel lines
D hyperbola
Points at distance d from a given line lie on two lines parallel to it, one on each side, because perpendicular distance to the original line stays constant.
Minimum distance from a point to a line is along
A parallel
B perpendicular
C angle bisector
D median
The shortest segment from a point to a line is the perpendicular from the point to the line. Any slant segment is longer than the perpendicular.
If slopes are m and m, lines are
A parallel
B perpendicular
C always intersect
D always vertical
Equal slopes mean same direction. If they are different lines, they never meet and are parallel; if constants also match, they are the same line.
Point-line position test uses sign of
A √(a²+b²)
B x1+y1
C ax1+by1+c
D a²+b²
Substitute the point in ax+by+c=0. The sign of ax1+by1+c tells which side of the line the point lies on, and zero means on the line.
Foot of perpendicular from P to a line is
A farthest point
B midpoint always
C origin always
D nearest point
The foot of the perpendicular is the point on the line where the perpendicular from P meets it. This perpendicular gives the shortest distance, so it is the nearest point.
The line y = x makes with x-axis an angle
A 30°
B 45°
C 60°
D 90°
For a line making angle θ with positive x-axis, slope m = tanθ. Here slope is 1, so tanθ = 1, giving θ = 45°.
For y = −x, the inclination angle is
A 45°
B 90°
C 135°
D 0°
Slope is −1, so tanθ = −1. Inclination θ is taken between 0° and 180°, giving θ = 135° for tanθ = −1.
Equation of a median needs
A vertex and midpoint
B two vertices only
C centroid only
D slope only
A median joins a vertex to the midpoint of the opposite side. So to write its equation, we use the vertex and that midpoint as two points.
Perpendicular bisector of a segment passes through
A segment endpoint
B origin only
C segment midpoint
D centroid only
A perpendicular bisector must pass through the midpoint of the segment and be perpendicular to the segment. This ensures equal distance from both endpoints.
Area method for collinearity checks area equals
A 1
B 0
C 2
D 3
Three points are collinear if the triangle area formed by them is zero. Using coordinate area formula, if computed area is 0, points lie on one line