Exercise 9.3 — Straight Lines

1. 9.3.14 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Reduce the following equations into slope-intercept form and find their slopes and y-intercepts: (i) x+7y=0, (ii) 6x+3y-5=0, (iii) y=0.

   Hint. Isolate y in each equation to match the y=mx+c form, then read off m and c directly.

   (i) x+7y=0 gives y=-x/7, so slope=-1/7, y-intercept=0. (ii) 6x+3y-5=0 gives y=-2x+5/3, so slope=-2, y-intercept=5/3. (iii) y=0 is already in this form with slope=0 and y-intercept=0.

   ✦ Working through each part gives: (i) slope -1/7, y-intercept 0. (ii) slope -2, y-intercept 5/3. (iii) slope 0, y-intercept 0.

2. 9.3.24 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Reduce the following equations into intercept form and find their intercepts on the axes: (i) 3x+2y-12=0, (ii) 4x-3y=6, (iii) 3y+2=0.

   Hint. Rearrange each equation into x/a+y/b=1 form by dividing through so the constant term becomes 1.

   (i) 3x+2y=12 gives x/4+y/6=1, so x-intercept=4, y-intercept=6. (ii) 4x-3y=6 gives x/(3/2)+y/(-2)=1, so x-intercept=3/2, y-intercept=-2. (iii) 3y=-2 gives y=-2/3, a line parallel to the x-axis with no x-intercept and y-intercept=-2/3.

   ✦ Working through each part gives: (i) x-intercept 4, y-intercept 6. (ii) x-intercept 3/2, y-intercept -2. (iii) no x-intercept (parallel to x-axis); y-intercept -2/3.

3. 9.3.33 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the distance of the point (-1,1) from the line 12(x+6) = 5(y-2).

   Hint. Expand the given equation into general form Ax+By+C=0 first, then apply the distance formula.

   Expanding: 12x+72=5y-10, giving 12x-5y+82=0. Distance from (-1,1): |12(-1)-5(1)+82|/sqrt(144+25) = |-12-5+82|/13 = 65/13 = 5.

   ✦ Working through each part gives: 5 units.

4. 9.3.44 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the points on the x-axis whose distances from the line x/3 + y/4 = 1 are 4 units.

   Hint. Convert the line to general form, then set the distance from a general point (x,0) equal to 4 and solve for x.

   The line x/3+y/4=1 becomes 4x+3y-12=0. For a point (x,0): |4x-12|/5=4, so |4x-12|=20, giving 4x-12=20 (x=8) or 4x-12=-20 (x=-2).

   ✦ Working through each part gives: (8, 0) and (-2, 0).

5. 9.3.54 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the distance between the parallel lines: (i) 15x+8y-34=0 and 15x+8y+31=0, (ii) l(x+y)+p=0 and l(x+y)-r=0.

   Hint. Apply the distance-between-parallel-lines formula directly in each case, being careful with the general form's A, B, C values in part (ii).

   (i) d=|(-34)-31|/sqrt(15^2+8^2)=65/sqrt(289)=65/17. (ii) Writing both as lx+ly+p=0 and lx+ly-r=0: A=l,B=l,C1=p,C2=-r, so d=|p-(-r)|/sqrt(l^2+l^2)=|p+r|/(l.sqrt2).

   ✦ Working through each part gives: (i) 65/17 units. (ii) |p+r|/(l.sqrt2) units.

6. 9.3.63 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the equation of the line parallel to the line 3x-4y+2=0 and passing through the point (-2,3).

   Hint. A line parallel to 3x-4y+2=0 has the same A and B coefficients, only C differs; find C from the given point.

   The parallel line has the form 3x-4y+k=0. Substituting (-2,3): 3(-2)-4(3)+k=0, giving -6-12+k=0, so k=18.

   ✦ Working through each part gives: 3x - 4y + 18 = 0.

7. 9.3.74 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the equation of the line perpendicular to the line x-7y+5=0 and having x-intercept 3.

   Hint. Find the perpendicular slope from the given line, then apply point-slope form through the point (3,0).

   Slope of x-7y+5=0 is 1/7, so the perpendicular slope is -7. Through (3,0): y-0=-7(x-3), giving y=-7x+21, i.e. 7x+y-21=0.

   ✦ Working through each part gives: 7x + y - 21 = 0.

8. 9.3.84 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the angle between the lines sqrt3.x + y = 1 and x + sqrt3.y = 1.

   Hint. Find both slopes, then apply the angle-between-two-lines formula.

   Slope of sqrt3.x+y=1 is -sqrt3. Slope of x+sqrt3.y=1 is -1/sqrt3. tan(theta)=|(-sqrt3-(-1/sqrt3))/(1+(-sqrt3)(-1/sqrt3))|=|(-2/sqrt3)/2|=1/sqrt3, giving theta=30 degrees.

   ✦ Working through each part gives: 30 degrees (the other angle between the lines is 150 degrees).

9. 9.3.94 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   The line through the points (h,3) and (4,1) intersects the line 7x-9y-19=0 at right angle. Find the value of h.

   Hint. Find the slope of the given line, take its perpendicular, and set it equal to the slope through the two given points.

   Slope of 7x-9y-19=0 is 7/9. Slope through (h,3),(4,1) is (1-3)/(4-h)=-2/(4-h). Perpendicularity: -2/(4-h)=-1, giving 14=9(4-h)=36-9h, so 9h=22, h=22/9.

   ✦ Working through each part gives: h = 22/9.

10. 9.3.104 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Prove that the line through the point (x1,y1) and parallel to the line Ax+By+C=0 is A(x-x1)+B(y-y1)=0.

    Hint. A line parallel to Ax+By+C=0 shares the same A and B; find the new constant from the condition that it passes through (x1,y1).

    A line parallel to Ax+By+C=0 has the form Ax+By+C'=0 for some constant C'. Since it passes through (x1,y1): Ax1+By1+C'=0, so C'=-(Ax1+By1). Substituting back: Ax+By-(Ax1+By1)=0, which rearranges to A(x-x1)+B(y-y1)=0.

    ✦ Working through each part gives: proved: A(x - x1) + B(y - y1) = 0.

11. 9.3.115 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Two lines passing through the point (2,3) intersect each other at an angle of 60 degrees. If the slope of one line is 2, find the equation of the other line.

    Hint. Substitute into the angle formula with m1=2 and theta=60 degrees, solve both sign cases for the unknown slope, then apply point-slope form.

    tan60=sqrt3=|(2-m)/(1+2m)|. Solving both cases and rationalising gives m=(5sqrt3-8)/11 or m=-(8+5sqrt3)/11. Using point-slope form through (2,3) with each slope and clearing denominators gives the two possible lines.

    ✦ Working through each part gives: (5sqrt3-8)x - 11y + (49-10sqrt3) = 0, or (8+5sqrt3)x + 11y - (49+10sqrt3) = 0.

12. 9.3.124 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the equation of the right bisector of the line segment joining the points (3,4) and (-1,2).

    Hint. The right bisector passes through the midpoint of the segment, with slope perpendicular to the segment.

    Midpoint of (3,4) and (-1,2) is (1,3). Slope of the segment is (2-4)/(-1-3)=1/2, so the perpendicular slope is -2. Through (1,3): y-3=-2(x-1), giving y-3=-2x+2, i.e. 2x+y-5=0.

    ✦ Working through each part gives: 2x + y - 5 = 0.

13. 9.3.134 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the coordinates of the foot of the perpendicular from the point (-1,3) to the line 3x-4y-16=0.

    Hint. Find the perpendicular line through the given point, then solve it simultaneously with the given line.

    Slope of 3x-4y-16=0 is 3/4, so the perpendicular slope is -4/3. The perpendicular through (-1,3): 4x+3y-5=0. Solving simultaneously with 3x-4y-16=0 (eliminate y by combining 3x-4y=16 and 4x+3y=5 appropriately) gives x=68/25, y=-49/25.

    ✦ Working through each part gives: (68/25, -49/25).

14. 9.3.144 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    The perpendicular from the origin to the line y=mx+c meets it at the point (-1,2). Find the values of m and c.

    Hint. Use the fact that (-1,2) lies on the line to get one equation, and that the segment from the origin to (-1,2) is perpendicular to the line to get a second.

    Since (-1,2) is on the line: 2=-m+c, so c=2+m. The segment from (0,0) to (-1,2) has slope -2; perpendicularity requires (-2)(m)=-1, giving m=1/2. Then c=2+1/2=5/2.

    ✦ Working through each part gives: m = 1/2, c = 5/2.

15. 9.3.156 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    If p and q are the lengths of perpendiculars from the origin to the lines x.cos(theta) - y.sin(theta) = k.cos(2theta) and x.sec(theta) + y.cosec(theta) = k, respectively, prove that p^2 + 4q^2 = k^2.

    Hint. Compute each perpendicular distance from the origin directly, using the Pythagorean identity to simplify the denominators.

    For the first line, distance from origin gives p=|k.cos2theta|, so p^2=k^2.cos^2(2theta). For the second line, sec^2theta+cosec^2theta simplifies to 1/(sin^2theta.cos^2theta)=4/sin^2(2theta), so q=|k|.sin(2theta)/2, giving 4q^2=k^2.sin^2(2theta). Adding: p^2+4q^2=k^2[cos^2(2theta)+sin^2(2theta)]=k^2.

    ✦ Working through each part gives: proved: p^2 + 4q^2 = k^2.

16. 9.3.165 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    In the triangle ABC with vertices A(2,3), B(4,-1), and C(1,2), find the equation and length of the altitude from the vertex A.

    Hint. The altitude from A is perpendicular to BC and passes through A; find its equation, then compute the distance from A to line BC for the length.

    Slope of BC: (2-(-1))/(1-4)=-1, so the altitude's slope is 1. Equation through A(2,3): y-3=1(x-2), giving x-y+1=0. Line BC (through B and C, slope -1): x+y-3=0. Length of altitude = distance from A(2,3) to x+y-3=0 = |2+3-3|/sqrt2 = 2/sqrt2 = sqrt2.

    ✦ Working through each part gives: altitude equation: x - y + 1 = 0. Length: sqrt2 units.

17. 9.3.174 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    If p is the length of the perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p^2 = 1/a^2 + 1/b^2.

    Hint. Write the intercept-form line in general form, apply the distance-from-origin formula, then take the reciprocal of p-squared.

    The line x/a+y/b=1 becomes bx+ay-ab=0. Distance from origin: p=|ab|/sqrt(a^2+b^2), so p^2=a^2b^2/(a^2+b^2). Then 1/p^2=(a^2+b^2)/(a^2b^2)=1/b^2+1/a^2.

    ✦ Working through each part gives: proved: 1/p^2 = 1/a^2 + 1/b^2.

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.