Miscellaneous Exercise — Straight Lines

1. 9.M.15 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the values of k for which the line (k-3)x - (4-k^2)y + k^2-7k+6 = 0 is (a) parallel to the x-axis, (b) parallel to the y-axis, (c) passing through the origin.

   Hint. Parallel to the x-axis needs the x-coefficient to vanish; parallel to the y-axis needs the y-coefficient to vanish; passing through the origin needs the constant term to vanish.

   (a) k-3=0 gives k=3 (and the y-coefficient 4-9=-5 is nonzero, so this is valid). (b) 4-k^2=0 gives k=2 or k=-2 (both keep the x-coefficient nonzero). (c) k^2-7k+6=0 factors as (k-1)(k-6)=0, giving k=1 or k=6.

   ✦ Working through each part gives: (a) k=3. (b) k=2 or k=-2. (c) k=1 or k=6.

2. 9.M.24 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the equations of the lines which cut off intercepts on the axes whose sum and product are 1 and -6, respectively.

   Hint. The two intercepts are roots of a quadratic built from their sum and product; solve it, then write the intercept form for each ordering.

   The intercepts a,b satisfy a+b=1, ab=-6, so they are roots of t^2-t-6=0, i.e. (t-3)(t+2)=0, giving {a,b}={3,-2}. With a=3,b=-2: x/3-y/2=1, i.e. 2x-3y-6=0. With a=-2,b=3: -x/2+y/3=1, i.e. 3x-2y+6=0.

   ✦ Working through each part gives: 2x - 3y - 6 = 0, or 3x - 2y + 6 = 0.

3. 9.M.34 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?

   Hint. Convert to general form, then apply the distance formula to a general point (0,y) on the y-axis.

   The line becomes 4x+3y-12=0. For (0,y): |3y-12|/5=4, so |3y-12|=20, giving 3y-12=20 (y=32/3) or 3y-12=-20 (y=-8/3).

   ✦ Working through each part gives: (0, 32/3) and (0, -8/3).

4. 9.M.45 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the perpendicular distance from the origin to the line joining the points (cos(theta), sin(theta)) and (cos(phi), sin(phi)).

   Hint. Find the equation of the line using sum-to-product identities to simplify the slope, then apply the distance-from-origin formula.

   Using sum-to-product identities, the slope simplifies to -cot((theta+phi)/2). Writing the line's equation and simplifying using the cosine-difference identity gives x.cos((theta+phi)/2)+y.sin((theta+phi)/2)=cos((phi-theta)/2), which is already in normal form. The distance from the origin is therefore |cos((phi-theta)/2)|.

   ✦ Working through each part gives: cos((theta-phi)/2).

5. 9.M.54 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the equation of the line parallel to the y-axis and drawn through the point of intersection of the lines x-7y+5=0 and 3x+y=0.

   Hint. Solve the two lines simultaneously to find the intersection point, then write a vertical line through that x-coordinate.

   From 3x+y=0, y=-3x. Substituting into x-7y+5=0: x-7(-3x)+5=0, giving 22x=-5, so x=-5/22. A line parallel to the y-axis through this point is x=-5/22.

   ✦ Working through each part gives: x = -5/22.

6. 9.M.64 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point where it meets the y-axis.

   Hint. Find where the given line meets the y-axis, find the perpendicular slope, then apply point-slope form.

   The line meets the y-axis at (0,6). Slope of x/4+y/6=1 is -3/2, so the perpendicular slope is 2/3. Through (0,6): y-6=(2/3)(x), giving 3y-18=2x, i.e. 2x-3y+18=0.

   ✦ Working through each part gives: 2x - 3y + 18 = 0.

7. 9.M.74 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the area of the triangle formed by the lines y-x=0, x+y=0, and x-k=0.

   Hint. Find the three pairwise intersection points, then apply the coordinate area formula.

   y=x meets x=k at (k,k). y=-x meets x=k at (k,-k). y=x meets y=-x at (0,0). Area = (1/2)|0(k-(-k))+k(-k-0)+k(0-k)| = (1/2)|-k^2-k^2| = k^2.

   ✦ Working through each part gives: k^2 square units.

8. 9.M.84 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the value of p so that the three lines 3x+y-2=0, px+2y-3=0, and 2x-y-3=0 may intersect at one point.

   Hint. Find the intersection of the two lines not involving p, then substitute that point into the third equation.

   Solving 3x+y=2 and 2x-y=3 (adding gives 5x=5, x=1, then y=-1): intersection is (1,-1). Substituting into px+2y-3=0: p(1)+2(-1)-3=0, giving p=5.

   ✦ Working through each part gives: p = 5.

9. 9.M.95 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   If three lines whose equations are y=m1x+c1, y=m2x+c2, and y=m3x+c3 are concurrent, then show that m1(c2-c3) + m2(c3-c1) + m3(c1-c2) = 0.

   Hint. Find the intersection point of the first two lines algebraically, then require that point to satisfy the third line's equation, and rearrange.

   Solving lines 1 and 2 gives x=(c2-c1)/(m1-m2) and y=(m1c2-m2c1)/(m1-m2). Substituting into y=m3x+c3 and multiplying through by (m1-m2) gives m1c2-m2c1=m3(c2-c1)+c3(m1-m2), which rearranges exactly to m1(c2-c3)+m2(c3-c1)+m3(c1-c2)=0.

   ✦ Working through each part gives: proved: m1(c2-c3) + m2(c3-c1) + m3(c1-c2) = 0.

10. 9.M.105 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the equation of the lines through the point (3,2) which make an angle of 45 degrees with the line x-2y=3.

    Hint. Find the slope of the given line, substitute into the angle formula with theta=45, solve both sign cases, then apply point-slope form.

    Slope of x-2y=3 is 1/2. |(m-1/2)/(1+m/2)|=1 gives m=3 (from the + case) or m=-1/3 (from the - case). Through (3,2): with m=3, y-2=3(x-3), i.e. 3x-y-7=0; with m=-1/3, y-2=-1/3(x-3), i.e. x+3y-9=0.

    ✦ Working through each part gives: 3x - y - 7 = 0, or x + 3y - 9 = 0.

11. 9.M.115 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the equation of the line passing through the point of intersection of the lines 4x+7y-3=0 and 2x-3y+1=0 that has equal intercepts on the axes.

    Hint. Find the intersection point, then consider both the equal-nonzero-intercept case (x+y=a) and the through-the-origin case (both intercepts trivially zero).

    Solving 4x+7y=3 and 2x-3y=-1 gives the intersection (1/13, 5/13). For a line x+y=a through this point: a=1/13+5/13=6/13, giving 13x+13y-6=0. For a line through the origin (both intercepts zero) and this point: slope=5, giving 5x-y=0.

    ✦ Working through each part gives: 5x - y = 0, or 13x + 13y - 6 = 0.

12. 9.M.125 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Show that the equation of the line passing through the origin and making an angle theta with the line y=mx+c is y/x = (m +- tan(theta)) / (1 -+ m.tan(theta)).

    Hint. Let the required line have slope m'; apply the angle formula between m and m', solving both sign cases, then write y=m'x since the line passes through the origin.

    Setting tan(theta)=(m'-m)/(1+m'm) and solving for m' gives m'=(m+tan theta)/(1-m.tan theta). Setting -tan(theta)=(m'-m)/(1+m'm) instead gives m'=(m-tan theta)/(1+m.tan theta). Combining both cases: m'=(m +- tan theta)/(1 -+ m.tan theta). Since the line passes through the origin, its equation is y=m'x, i.e. y/x=m'.

    ✦ Working through each part gives: proved: y/x = (m +- tan(theta)) / (1 -+ m.tan(theta)).

13. 9.M.134 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    In what ratio is the line joining (-1,1) and (5,7) divided by the line x+y=4?

    Hint. Let the dividing point (using the section formula with ratio k:1) satisfy x+y=4, and solve for k.

    The point dividing (-1,1),(5,7) in ratio k:1 is ((5k-1)/(k+1),(7k+1)/(k+1)). Substituting into x+y=4: (5k-1+7k+1)/(k+1)=4, giving 12k=4(k+1)=4k+4, so 8k=4, k=1/2.

    ✦ Working through each part gives: the ratio is 1:2.

14. 9.M.145 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the distance of the line 4x+7y+5=0 from the point (1,2) along the line 2x-y=0.

    Hint. Find where the line 2x-y=0 (which passes through (1,2)) meets 4x+7y+5=0, then compute the ordinary distance between (1,2) and that intersection point.

    From 2x-y=0, y=2x. Substituting into 4x+7y+5=0: 4x+14x+5=0, giving x=-5/18, y=-5/9. Distance from (1,2) to (-5/18,-5/9): differences are 23/18 and 46/18, so distance = (1/18).sqrt(23^2+46^2) = (1/18).sqrt(2645) = (1/18)(23.sqrt5) = 23.sqrt5/18.

    ✦ Working through each part gives: 23.sqrt5 / 18.

15. 9.M.155 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the direction in which a straight line must be drawn through the point (-1,2) so that its point of intersection with the line x+y=4 may be at a distance of 3 units from this point.

    Hint. Parametrise points on the required line at distance r from (-1,2) using its inclination theta, require the point at r=3 to satisfy x+y=4, and solve for theta.

    A point at distance r along inclination theta from (-1,2) is (-1+r.cos theta, 2+r.sin theta). At r=3, substituting into x+y=4: 1+3(cos theta+sin theta)=4, so cos theta+sin theta=1, i.e. sqrt2.sin(theta+45)=1, giving theta+45=45 or theta+45=135, so theta=0 or theta=90.

    ✦ Working through each part gives: the line must be drawn horizontally (0 degrees) or vertically (90 degrees) from (-1,2).

16. 9.M.165 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    The hypotenuse of a right angled triangle has its ends at the points (1,3) and (-4,1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

    Hint. The right-angle vertex, for legs parallel to the axes, sits where a horizontal line through one endpoint meets a vertical line through the other -- there are two such configurations.

    One configuration has the right-angle vertex at (1,1): the legs are x=1 (through (1,3) and (1,1)) and y=1 (through (-4,1) and (1,1)). The other has the vertex at (-4,3): the legs are x=-4 (through (-4,1) and (-4,3)) and y=3 (through (1,3) and (-4,3)).

    ✦ Working through each part gives: x = 1 and y = 1 (one triangle), or x = -4 and y = 3 (the other possible triangle).

17. 9.M.174 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the image of the point (3,8) with respect to the line x+3y=7, assuming the line to be a plane mirror.

    Hint. Apply the reflection formula, which moves the point twice the signed perpendicular distance along the line's normal direction.

    With A=1,B=3,C=-7: Ax1+By1+C=3+24-7=20, and A^2+B^2=10. Image x = 3 - 2(1)(20)/10 = 3-4=-1. Image y = 8 - 2(3)(20)/10 = 8-12=-4.

    ✦ Working through each part gives: (-1, -4).

18. 9.M.186 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    If the lines y=3x+1 and 2y=x+3 are equally inclined to the line y=mx+4, find the value of m.

    Hint. Set the tangent of the angle from the mirror line to the first line equal to the negative of the tangent of the angle to the second line (mirror symmetry), then solve the resulting quadratic in m.

    With slopes 3 and 1/2 for the two given lines: (3-m)/(1+3m) = -(1/2-m)/(1+m/2). Cross-multiplying and simplifying leads to 7m^2-2m-7=0, giving m=(2 +- sqrt(4+196))/14=(1 +- 5.sqrt2)/7.

    ✦ Working through each part gives: m = (1+5.sqrt2)/7 or m = (1-5.sqrt2)/7.

19. 9.M.195 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    If the sum of the perpendicular distances of a variable point P(x,y) from the lines x+y-5=0 and 3x-2y+7=0 is always 10, show that P must move on a line.

    Hint. Assuming a region where both expressions inside the absolute values keep a consistent sign, drop the absolute value bars and simplify -- the result is linear in x and y.

    Assuming x+y-5>=0 and 3x-2y+7>=0 in the relevant region: (x+y-5)/sqrt2 + (3x-2y+7)/sqrt13 = 10. Multiplying through by sqrt26 and collecting terms gives an equation of the form Ax+By+C=0 with constant coefficients -- a straight line. The same reasoning applies (with different signs) in each of the other three sign regions, so within any single region P's path is confined to a straight line.

    ✦ Working through each part gives: proved: within any region of consistent sign for the two expressions, P's locus reduces to a linear equation Ax+By+C=0, i.e. a straight line.

20. 9.M.204 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Find the equation of the line which is equidistant from the parallel lines 9x+6y-7=0 and 3x+2y+6=0.

    Hint. Rewrite both lines with matching A and B coefficients, then take the equidistant line's constant term as the average of the two.

    Scaling the first line by dividing by 3: 3x+2y-7/3=0. The equidistant line has the same A,B with C midway between -7/3 and 6: C=(-7/3+6)/2=11/6. So the line is 3x+2y+11/6=0, i.e. 18x+12y+11=0.

    ✦ Working through each part gives: 18x + 12y + 11 = 0.

21. 9.M.215 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    A ray of light passing through the point (1,2) reflects on the x-axis at point A, and the reflected ray passes through the point (5,3). Find the coordinates of A.

    Hint. The reflected ray is the straight-line continuation from the mirror image of (1,2) in the x-axis, through A, to (5,3) -- so A is where the line joining that image point and (5,3) crosses the x-axis.

    The image of (1,2) reflected in the x-axis is (1,-2). The line through (1,-2) and (5,3) has slope 5/4: y+2=(5/4)(x-1). Setting y=0: 2=(5/4)(x-1), giving x-1=8/5, x=13/5.

    ✦ Working through each part gives: a = (13/5, 0).

22. 9.M.226 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    Prove that the product of the lengths of the perpendiculars drawn from the points (sqrt(a^2-b^2),0) and (-sqrt(a^2-b^2),0) to the line (x/a)cos(theta)+(y/b)sin(theta)=1 is b^2.

    Hint. Write both perpendicular distances using the distance formula, then multiply them together -- the resulting expression simplifies dramatically once a common factor cancels between numerator and denominator.

    With A=cos(theta)/a, B=sin(theta)/b, C=-1, and e=(cos theta/a).sqrt(a^2-b^2): the two distances are |e-1|/D and |e+1|/D where D=sqrt(A^2+B^2). Their product is |e^2-1|/D^2. Computing e^2-1 = -sin^2theta - (b^2/a^2)cos^2theta, so |e^2-1| = sin^2theta+(b^2/a^2)cos^2theta = [a^2sin^2theta+b^2cos^2theta]/a^2. Also D^2 = [b^2cos^2theta+a^2sin^2theta]/(a^2b^2). Dividing, the [a^2sin^2theta+b^2cos^2theta] terms cancel exactly, leaving the product equal to b^2.

    ✦ Working through each part gives: proved: the product of the two perpendicular distances equals b^2.

23. 9.M.235 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    A person standing at the junction (crossing) of two straight paths represented by the equations 2x-3y+4=0 and 3x+4y-5=0 wants to reach the path whose equation is 6x-7y+8=0 in the least time. Find the equation of the path that he should follow.

    Hint. The least-time path is the perpendicular from the junction point to the target path; find the junction (intersection of the two given paths) first, then build the perpendicular line through it.

    Solving 2x-3y+4=0 and 3x+4y-5=0 simultaneously gives the junction point (-1/17, 22/17). The target line 6x-7y+8=0 has slope 6/7, so the perpendicular path has slope -7/6. Through (-1/17,22/17) with slope -7/6, clearing denominators gives 119x+102y-125=0.

    ✦ Working through each part gives: 119x + 102y - 125 = 0.

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.