Write the equations for the x-axis and the y-axis.
Hint. Every point on the x-axis has y=0; every point on the y-axis has x=0.
The x-axis consists of all points with y-coordinate 0, so its equation is y=0. The y-axis consists of all points with x-coordinate 0, so its equation is x=0.
✦ Working through each part gives: x-axis: y = 0. y-axis: x = 0.
Find the equation of the line passing through the point (-4,3) with slope 1/2.
Hint. Apply the point-slope form directly.
Using y-y0=m(x-x0) with (x0,y0)=(-4,3), m=1/2: y-3=(1/2)(x+4). Multiplying by 2: 2y-6=x+4, giving x-2y+10=0.
✦ Working through each part gives: x - 2y + 10 = 0.
Find the equation of the line passing through (0,0) with slope m.
Hint. Apply point-slope form through the origin.
Using y-0=m(x-0): y=mx, i.e. mx-y=0.
✦ Working through each part gives: y = mx, i.e. mx - y = 0.
Find the equation of the line passing through (2, 2sqrt3) and inclined with the x-axis at an angle of 75 degrees.
Hint. Find tan(75) using the tan(45+30) addition formula, then apply point-slope form.
tan75 = tan(45+30) = (1+1/sqrt3)/(1-1/sqrt3) = (sqrt3+1)/(sqrt3-1), which rationalises to 2+sqrt3. Using point-slope with (2,2sqrt3): y-2sqrt3=(2+sqrt3)(x-2). Expanding: y=(2+sqrt3)x-2(2+sqrt3)+2sqrt3=(2+sqrt3)x-4-2sqrt3+2sqrt3=(2+sqrt3)x-4.
✦ Working through each part gives: (2+sqrt3)x - y - 4 = 0.
Find the equation of the line intersecting the x-axis at a distance of 3 units to the left of the origin, with slope -2.
Hint. The condition describes the point (-3,0); apply point-slope form there.
The line passes through (-3,0) with slope -2: y-0=-2(x-(-3))=-2(x+3), giving y=-2x-6, i.e. 2x+y+6=0.
✦ Working through each part gives: 2x + y + 6 = 0.
Find the equation of the line intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30 degrees with the positive direction of the x-axis.
Hint. The condition describes the point (0,2) with slope tan(30); apply point-slope form.
The line passes through (0,2) with slope tan30=1/sqrt3: y-2=(1/sqrt3)x. Multiplying by sqrt3: sqrt3.y-2sqrt3=x, giving x-sqrt3.y+2sqrt3=0.
✦ Working through each part gives: x - sqrt3.y + 2sqrt3 = 0.
Find the equation of the line passing through the points (-1,1) and (2,-4).
Hint. Apply the two-point form directly.
Slope = (-4-1)/(2-(-1)) = -5/3. Using point-slope with (-1,1): y-1=(-5/3)(x+1). Multiplying by 3: 3y-3=-5x-5, giving 5x+3y+2=0.
✦ Working through each part gives: 5x + 3y + 2 = 0.
The vertices of triangle PQR are P(2,1), Q(-2,3), and R(4,5). Find the equation of the median through the vertex R.
Hint. The median from R goes to the midpoint of the opposite side PQ; find that midpoint, then apply two-point form.
Midpoint of PQ = ((2-2)/2,(1+3)/2) = (0,2). Slope of the median through R(4,5) and (0,2): (5-2)/(4-0)=3/4. Equation: y-2=(3/4)(x-0), i.e. 4y-8=3x, giving 3x-4y+8=0.
✦ Working through each part gives: 3x - 4y + 8 = 0.
Find the equation of the line passing through (-3,5) and perpendicular to the line through the points (2,5) and (-3,6).
Hint. Find the slope of the given line, take its negative reciprocal, then apply point-slope form.
Slope of the line through (2,5),(-3,6): (6-5)/(-3-2)=-1/5. Perpendicular slope = 5. Equation through (-3,5): y-5=5(x+3), giving y-5=5x+15, i.e. 5x-y+20=0.
✦ Working through each part gives: 5x - y + 20 = 0.
A line perpendicular to the line segment joining the points (1,0) and (2,3) divides it in the ratio 1:n. Find the equation of the line.
Hint. Find the slope of the given segment and its perpendicular slope, find the dividing point using the section formula in terms of n, then apply point-slope form.
Slope of segment (1,0)-(2,3) is 3, so the perpendicular slope is -1/3. The point dividing (1,0),(2,3) in ratio 1:n is ((2+n)/(1+n), 3/(1+n)). Using point-slope form with this point and slope -1/3, then multiplying through by 3(1+n) and simplifying gives x+3y-11+n(x+3y-1)=0.
✦ Working through each part gives: x + 3y - 11 + n(x + 3y - 1) = 0.
Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).
Hint. Equal intercepts a=b turns the intercept form into x+y=a; substitute the given point to find a.
With equal intercepts a=b, the intercept form x/a+y/a=1 becomes x+y=a. Substituting (2,3): 2+3=a=5. So the equation is x+y=5.
✦ Working through each part gives: x + y - 5 = 0.
Find the equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9.
Hint. Write a+b=9, substitute the point into the intercept form, and solve the resulting quadratic in a.
With b=9-a and the point (2,2) on x/a+y/b=1: 2/a+2/(9-a)=1. Multiplying through by a(9-a): 2(9-a)+2a=a(9-a), giving 18=9a-a^2, i.e. a^2-9a+18=0, so (a-6)(a-3)=0, a=6 or a=3.
✦ Working through each part gives: x+2y-6=0 (from a=6,b=3) or 2x+y-6=0 (from a=3,b=6).
Find the equation of the line through the point (0,2) making an angle 2(pi)/3 with the positive x-axis. Also find the equation of the line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Hint. Compute the slope from tan(2pi/3), then write both parallel lines using point-slope form through their respective y-axis points.
Slope = tan(2pi/3) = tan(120 degrees) = -sqrt3. Through (0,2): y-2=-sqrt3.x, giving sqrt3.x+y-2=0. The parallel line through (0,-2) (2 units below the origin): y+2=-sqrt3.x, giving sqrt3.x+y+2=0.
✦ Working through each part gives: sqrt3.x + y - 2 = 0, and the parallel line sqrt3.x + y + 2 = 0.
The perpendicular from the origin to a line meets it at the point (-2,9). Find the equation of the line.
Hint. The line's slope is the negative reciprocal of the slope from the origin to (-2,9); apply point-slope form there.
Slope of the segment from (0,0) to (-2,9) is 9/(-2)=-9/2, so the line's slope is the negative reciprocal, 2/9. Through (-2,9): y-9=(2/9)(x+2). Multiplying by 9: 9y-81=2x+4, giving 2x-9y+85=0.
✦ Working through each part gives: 2x - 9y + 85 = 0.
The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, L=124.942 when C=20 and L=125.134 when C=110. Express L in terms of C.
Hint. Treat (C,L) as two points and apply the two-point form of a line, with C playing the role of x and L the role of y.
Slope = (125.134-124.942)/(110-20) = 0.192/90. Using point-slope through (20,124.942): L-124.942 = (0.192/90)(C-20).
✦ Working through each part gives: l = 124.942 + (0.192/90)(C - 20), approximately L = 124.942 + 0.00213(C - 20).
The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?
Hint. Treat demand as a linear function of price using the two given (price, demand) points, then substitute price=17.
Treating demand D as linear in price P, using points (14,980) and (16,1220): slope=(1220-980)/(16-14)=120. So D-980=120(P-14). At P=17: D=980+120(3)=980+360=1340.
✦ Working through each part gives: 1340 litres.
P(a,b) is the midpoint of a line segment between the axes. Show that the equation of the line is x/a + y/b = 2.
Hint. Let the axis intercepts be (p,0) and (0,q); their midpoint being (a,b) determines p and q in terms of a and b.
If the intercepts are (p,0) and (0,q), their midpoint is (p/2,q/2)=(a,b), so p=2a, q=2b. The intercept form x/p+y/q=1 becomes x/(2a)+y/(2b)=1, which simplifies (multiplying by 2) to x/a+y/b=2.
✦ Working through each part gives: proved: x/a + y/b = 2.
Point R(h,k) divides a line segment between the axes in the ratio 1:2. Find the equation of the line.
Hint. Let the intercepts be (p,0) and (0,q); use the section formula to express p and q in terms of h and k, then substitute into the intercept form.
With intercepts (p,0) and (0,q), R divides the segment from (p,0) to (0,q) in ratio 1:2, so R=(2p/3, q/3). Thus h=2p/3 (p=3h/2) and k=q/3 (q=3k). Substituting into x/p+y/q=1: x/(3h/2)+y/(3k)=1, i.e. 2x/h+y/k=3.
✦ Working through each part gives: 2x/h + y/k = 3.
By using the concept of equation of a line, prove that the three points (3,0), (-2,-2), and (8,2) are collinear.
Hint. Find the equation of the line through two of the points, then check whether the third point satisfies it.
Slope through (3,0),(-2,-2): (-2-0)/(-2-3)=2/5. Equation: y-0=(2/5)(x-3), i.e. 5y=2x-6, giving 2x-5y-6=0. Checking (8,2): 2(8)-5(2)-6=16-10-6=0, so it satisfies the equation.
✦ Working through each part gives: proved: (8,2) lies on the line through (3,0) and (-2,-2), so the three points are collinear.
Solutions written by the tuition.in editorial team and checked against the NCERT Class 11 Mathematics textbook, Reprint 2026-27 (kemh109.pdf) — Exercise 9.1 (11 questions), Exercise 9.2 (19 questions), Exercise 9.3 (17 questions), plus the chapter's Miscellaneous Exercise (23 questions), 70 questions total, the largest solutions file in this project to date. Confirmed against the CBSE curriculum PDF that Normal Form and 'general equation of a line' as a full topic are formative-only — Normal Form is completely absent from the current book (no formula, no example, no exercise question anywhere), while general form gets exactly one defining sentence before the distance-formula section, even though Exercise 9.3's own Q1-Q2 require converting general form into the other named forms. Several garbled stacked-fraction and trigonometric exercise questions (Ex 9.2 Q4/Q13; Ex 9.3 Q4/Q6/Q8/Q9/Q15; Misc Q3/Q6/Q12/Q22) were cross-checked via 300dpi page renders. Several multi-step algebraic results (Misc Q9, Q12, Q22) were verified symbolically term-by-term rather than by a single numeric spot check. Questions are referenced from the NCERT textbook for identification.
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