For the line 2x + 3y = 12, what are the x- and y-intercepts respectively?
A 4 and 6
B −6 and 4
C 6 and 4
D 6 and −4
Put y=0 gives 2x=12 so x=6. Put x=0 gives 3y=12 so y=4. Intercepts mean where the line cuts axes.
The intercept form is x/5 − y/2 = 1. What is its general form?
A 2x−5y−10=0
B 2x+5y−10=0
C 5x−2y−10=0
D 2x−5y+10=0
Multiply by 10: 2x−5y=102x−5y=10. Bring all terms to one side: 2x−5y−10=02x−5y−10=0. This represents the same line in standard general form.
A line passes through (1, −2) and has x-intercept 4. Which equation matches the line?
A 2x+3y−8=0
B 3x−2y−8=0
C 2x−3y+8=0
D 2x−3y−8=0
x-intercept 4 means point (4,0) lies on the line. Slope from (1,−2) to (4,0) is 2/3. Equation becomes 2x−3y−8=02x−3y−8=0.
A line passes through (2, 3) and also cuts the y-axis at −1. What is its equation?
A y=2x−1
B y=−2x−1
C y=x−1
D y=2x+1
The y-intercept −1 gives point (0,−1). Slope between (0,−1) and (2,3) is (3+1)/2=2(3+1)/2=2. So equation is y=2x−1y=2x−1.
Compare 3x + 4y + 1 = 0 and 6x + 8y − 5 = 0. What is their relationship?
A Perpendicular lines
B Same line
C Parallel lines
D Intersecting acute
Coefficients of x and y are proportional: 3:4 equals 6:8. But constants are not proportional (1 and −5), so lines are distinct but parallel.
For the line x + 2y − 7 = 0, which line through (1, 1) is perpendicular to it?
A x+2y−3=0
B 2x−y−1=0
C y=−2x+1
D x−2y−1=0
Given line slope is −A/B=−1/2−A/B=−1/2. Perpendicular slope is 2. Through (1,1): y−1=2(x−1)y−1=2(x−1) gives 2x−y−1=02x−y−1=0.
Two lines have slopes 1/2 and −2. What is the angle between them?
A 90°
B 0°
C 45°
D 135°
Perpendicular lines satisfy m1m2=−1m1m2=−1. Here (1/2)×(−2)=−1(1/2)×(−2)=−1. Therefore the lines meet at a right angle, so angle is 90°.
Lines 2x − y = 0 and x + 2y = 0 intersect. What is the acute angle between them?
A 0°
B 30°
C 90°
D 60°
First line gives y=2xy=2x slope 2. Second gives y=−x/2y=−x/2 slope −1/2. Product is −1, so the lines are perpendicular, giving 90°.
If one line has slope 1 and another has slope 2, what is the value of tanθ between them?
A 3
B 0
C √3
D 1/3
tanθ=∣m1−m21+m1m2∣=∣1−21+2∣=1/3tanθ=1+m1m2m1−m2=1+21−2=1/3. This gives the tangent of the acute angle.
Two non-vertical lines have equal slopes but different intercepts. What is the angle between them?
A 45°
B 0°
C 90°
D 180°
Equal slopes mean same direction, so lines do not meet (unless coincident). The smaller angle between their directions is 0°, indicating parallel distinct lines.
A line has direction ratios (2, −3). What is its slope?
A 3/2
B −2/3
C −3/2
D 2/3
Slope m=Δy/Δxm=Δy/Δx. With direction ratios (a,b) = (2,−3), slope is b/a=−3/2b/a=−3/2. It means y decreases as x increases.
In normal form x cosα + y sinα = p, if α = 0, the equation becomes
A x = p
B y = p
C y = mx
D Ax+By=0
cos0=1 and sin0=0, so the equation becomes x=px=p. That represents a vertical line at distance p from the origin along the x-direction.
The normal form is x cos90° + y sin90° = 5. What is the line?
A x = 5
B y = −5
C y = 5
D x = −5
cos90=0 and sin90=1, so equation becomes 0⋅x+1⋅y=50⋅x+1⋅y=5, giving y=5y=5. This is a horizontal line.
For 5x + 12y + 13 = 0, the perpendicular distance from origin equals
A 1
B 13
C 13/5
D 5/13
Distance from origin to Ax+By+C=0Ax+By+C=0 is ∣C∣/A2+B2∣C∣/A2+B2. Here ∣13∣/25+144=13/13=1∣13∣/25+144=13/13=1. So distance is 1 unit.
Lines x − y = 0 and x + y = 0 have angle bisectors. One bisector is
A y = x
B x = 0
C x + y = 0
D x − y = 0
Angle bisectors satisfy L1a12+b12=±L2a22+b22a12+b12L1=±a22+b22L2. Here it gives x=0x=0 and y=0y=0, the coordinate axes.
For lines 2x + y = 0 and x − 2y = 0, one angle bisector is
A x−3y=0
B 3x+y=0
C x+3y=0
D x+y=0
Since both have equal normal lengths, bisectors satisfy 2x+y=±(x−2y)2x+y=±(x−2y). Using plus gives 2x+y=x−2y⇒x+3y=02x+y=x−2y⇒x+3y=0, a correct bisector line.
The point (3, −1) lies on exactly which line?
A x+y−2=0
B x−y−2=0
C 2x+y−4=0
D x+2y−2=0
Substitute (3,−1). For option A: 3 + (−1) − 2 = 0, so it satisfies. The other equations give nonzero values, so they do not pass through the point.
Find the perpendicular distance from (2, −3) to the line 4x − 3y + 6 = 0
A 5/23
B 23/25
C 25/23
D 23/5
Use distance =∣ax1+by1+c∣a2+b2=a2+b2∣ax1+by1+c∣. Value is ∣4(2)−3(−3)+6∣/5=∣8+9+6∣/5=23/5∣4(2)−3(−3)+6∣/5=∣8+9+6∣/5=23/5.
For line ax + by + c = 0, if ax1 + by1 + c > 0, the point lies on
A Opposite side
B On the line
C Normal side
D Cannot decide
The sign of ax1+by1+cax1+by1+c tells the side relative to the normal vector (a,b). Positive means the point is on the side towards which the normal points.
Find the distance between 3x − 4y + 5 = 0 and 3x − 4y − 15 = 0
A 4
B 5
C 20
D 1/4
Distance between parallel lines ax+by+c1=0ax+by+c1=0 and ax+by+c2=0ax+by+c2=0 is ∣c1−c2∣/a2+b2∣c1−c2∣/a2+b2. Here 20/5=420/5=4.
A line is parallel to 2x − 3y + 6 = 0 and passes through origin. Its equation is
A 2x+3y=0
B 2x−3y=0
C 3x−2y=0
D 2x−3y+6=0
Parallel lines keep the same x and y coefficients. Passing through origin means constant term becomes 0, giving 2x−3y=02x−3y=0.
Lines x + 2y = 5 and 3x − y = 4 intersect at
A (11/7,13/7)
B (2,1)
C (13/7,11/7)
D (1,2)
Solve simultaneously: from first x=5−2yx=5−2y. Substitute into second: 3(5−2y)−y=4⇒15−7y=4⇒y=11/73(5−2y)−y=4⇒15−7y=4⇒y=11/7. Then x=13/7x=13/7.
Point P(1,2) divides A(−2,5) and B(4,−1) internally. Ratio AP:PB is
A 1:1
B 2:1
C 1:2
D 3:1
Using section formula with ratio m:n, x-coordinate gives (4m−2n)/(m+n)=1⇒m=n(4m−2n)/(m+n)=1⇒m=n. This also satisfies y-coordinate. So P is the midpoint, ratio 1:1.
Centroid of triangle with vertices (0,0), (6,0), (0,3) is
A (3,1)
B (2,1)
C (2,0)
D (1,1)
Centroid is average of coordinates: (0+6+03,0+0+33)=(2,1)(30+6+0,30+0+3)=(2,1). It is the balance point where medians meet.
The median from (0,0) to side joining (6,0) and (0,3) has equation
A 2x−y=0
B x+2y=0
C x−2y=0
D x−y=0
Midpoint of (6,0) and (0,3) is (3, 3/2). Line through (0,0) and (3,3/2) has slope 1/2, so y=x/2⇒x−2y=0y=x/2⇒x−2y=0.
Perpendicular bisector of segment joining (1,2) and (5,2) is
A x = 3
B y = 3
C y = 2
D x = 2
The segment is horizontal (same y). Midpoint is (3,2). A perpendicular to a horizontal line is vertical, so perpendicular bisector is x = 3.
Perpendicular bisector of segment joining (2,1) and (2,7) is
A x = 4
B y = 2
C x = 2
D y = 4
The segment is vertical (same x). Midpoint is (2,4). Perpendicular to a vertical line is horizontal, so the perpendicular bisector is y = 4.
Distance between points (−1,2) and (3,5) equals
A √7
B 7
C 5
D 4
Distance =(3+1)2+(5−2)2=42+32=25=5=(3+1)2+(5−2)2=42+32=25=5. This is a direct application of the distance formula.
A line passes through (1,1) and makes 45° with the positive x-axis. Its equation is
A y = −x
B y = x
C x = 1
D y = 1
Inclination 45° means slope m=tan45=1m=tan45=1. Through (1,1): y−1=1(x−1)⇒y=xy−1=1(x−1)⇒y=x. This line passes through origin and (1,1).
A line makes 135° with x-axis and passes through (2,0). Which equation fits?
A x+y−2=0
B x−y−2=0
C x+y+2=0
D x−y+2=0
Slope m=tan135=−1m=tan135=−1. Using point (2,0): y−0=−1(x−2)⇒y=−x+2⇒x+y−2=0y−0=−1(x−2)⇒y=−x+2⇒x+y−2=0.
Write 2x + 3y = 6 in intercept form
A x/2 + y/3 =1
B x/6 + y/6 =1
C x/3 − y/2 =1
D x/3 + y/2 =1
x-intercept: set y=0 gives x=3. y-intercept: set x=0 gives y=2. Intercept form becomes x/3+y/2=1x/3+y/2=1.
Parametric line x = 1 + 2t, y = 3 − t has symmetric form
A (x−1)/−2=(y−3)/1
B (x−3)/2=(y−1)/−1
C (x−1)/2=(y−3)/−1
D (x−1)/2=(y+3)/−1
From x and y, point is (1,3) and direction ratios are (2,−1). Symmetric form is (x−1)/2=(y−3)/(−1)(x−1)/2=(y−3)/(−1), describing the same line neatly.
A family of lines through (2, −1) with variable slope m is
A y+1=m(x−2)
B y−1=m(x−2)
C y+2=m(x+1)
D y=m(x−2)
Point-slope form is y−y1=m(x−x1)y−y1=m(x−x1). With (2,−1), it becomes y+1=m(x−2)y+1=m(x−2). Changing m gives all possible lines through that point.
When are two lines a1x+b1y+c1=0 and a2x+b2y+c2=0 coincident?
A Only slopes equal
B All ratios equal
C Only constants equal
D Product equals −1
Lines are coincident when a1a2=b1b2=c1c2a2a1=b2b1=c2c1. Then one equation is just a multiple of the other, representing the same line.
Compare 2x + 3y − 6 = 0 and 4x + 6y − 12 = 0. They are
A Parallel distinct
B Perpendicular
C Same line
D Intersecting lines
The second equation is exactly 2 times the first. Multiplying a line equation by a nonzero constant does not change the line, so they are coincident.
Compare x + 2y − 3 = 0 and 2x + 4y − 7 = 0. They are
A Parallel distinct
B Same line
C Perpendicular
D Intersecting 90°
Coefficients are proportional (x and y parts match ratio 1:2), but constants are not proportional (−3 and −7). Hence lines have same slope but are different, so parallel.
Slopes are 1 and −2. What is tanθ between the two lines?
A 3
B 1/3
C √3
D 0
tanθ=∣1−(−2)1+1⋅(−2)∣=∣3−1∣=3tanθ=1+1⋅(−2)1−(−2)=−13=3. This gives the tangent of the acute angle between the lines.
Check if x − √3y + 2 = 0 and √3x + y − 1 = 0 are perpendicular
A No, parallel
B Yes, perpendicular
C Same line
D Cannot decide
First line slope m1=−A/B=−1/(−3)=1/3m1=−A/B=−1/(−3)=1/3. Second slope m2=−3/1=−3m2=−3/1=−3. Product m1m2=−1m1m2=−1, so lines are perpendicular.
Distance from origin to 12x − 5y + 60 = 0 equals
A 13/60
B 60/169
C 60/13
D 13
Distance from (0,0) to ax+by+c=0ax+by+c=0 is ∣c∣/a2+b2∣c∣/a2+b2. Here ∣60∣/144+25=60/13∣60∣/144+25=60/13. This is the shortest distance.
A point is equidistant from lines y = 2 and y = 8. It must lie on
A x = 5
B y = 3
C y = 10
D y = 5
For two parallel horizontal lines, points equidistant from both lie on the middle line. The midpoint between y=2 and y=8 is y=(2+8)/2=5.
A point is equidistant from lines x = −1 and x = 7. It must lie on
A x = 3
B y = 3
C x = 4
D x = 0
These are vertical parallel lines. The set of points equidistant from them lies on the vertical midline at x = (−1+7)/2 = 3.
Points whose distance from y = 0 is 4 form
A One line
B One circle
C Two lines
D No locus
Distance from y=0 (x-axis) being 4 means y is 4 units above or below the x-axis. So the locus is y=4 and y=−4, two parallel lines.
If a line equation is multiplied by 5, the point-to-line distance becomes
A 5 times
B No change
C 1/5 times
D Becomes zero
Multiplying ax+by+c=0ax+by+c=0 by a nonzero constant multiplies numerator and denominator of distance formula equally, so the ratio stays the same. Hence distance remains unchanged.
In distance formula ∣ax1+by1+c∣/√(a2+b2)∣ax1+by1+c∣/√(a2+b2), √(a²+b²) represents
A Normal length
B Line intercept
C Slope value
D Angle measure
(a,b) is the normal vector to the line. √(a²+b²) is its length. Dividing by it converts the expression value into actual perpendicular distance.
A line is 3 units above the x-axis. Which equation represents it?
A y = −3
B x = 3
C y = 3
D x = −3
“Above x-axis” means positive y-value. A horizontal line at constant y=3 is exactly 3 units above the x-axis for every x.
A line passes through the intersection of x+y−4=0 and x−y−2=0 and also passes through origin. It is
A x−3y=0
B 3x−y=0
C x+3y=0
D x−y=0
Any line through intersection is L1+λL2=0L1+λL2=0. For origin: −4+λ(−2)=0⇒λ=−2−4+λ(−2)=0⇒λ=−2. Substitute gives (x+y−4)−2(x−y−2)=0⇒x−3y=0(x+y−4)−2(x−y−2)=0⇒x−3y=0.
A line is at distance 5 from origin and its normal makes 60° with x-axis. Its equation is
A √3x+y=10
B x−√3y=10
C x+√3y=5
D x+√3y=10
Normal form is xcosα+ysinα=pxcosα+ysinα=p. With α=60∘α=60∘, p=5p=5: x2+3y2=52x+23y=5. Multiply by 2 gives x+3y=10x+3y=10.