Exercise 9.1 — Straight Lines

1. 9.1.14 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Draw a quadrilateral in the Cartesian plane, whose vertices are (-4,5), (0,7), (5,-5), and (-4,-2). Also find its area.

   Hint. Split the quadrilateral along one diagonal into two triangles and sum their areas using the coordinate area formula.

   Using diagonal AC from A(-4,5) to C(5,-5): Area(ABC) with B(0,7) = (1/2)|(-4)(7-(-5))+0((-5)-5)+5(5-7)| = (1/2)|-48+0-10| = 29. Area(ACD) with D(-4,-2) = (1/2)|(-4)(-5-(-2))+5(-2-5)+(-4)(5-(-5))| = (1/2)|12-35-40| = 31.5. Total area = 29+31.5 = 60.5.

   ✦ Working through each part gives: area = 60.5 square units.

2. 9.1.24 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   The base of an equilateral triangle with side 2a lies along the y-axis such that the midpoint of the base is at the origin. Find the vertices of the triangle.

   Hint. Place the base endpoints symmetrically on the y-axis, then use the equilateral triangle's height formula to place the third vertex on the x-axis.

   Since the base has length 2a and its midpoint is the origin, the base endpoints are (0,a) and (0,-a). The height of an equilateral triangle with side 2a is sqrt((2a)^2-a^2)=a.sqrt3, so the third vertex lies on the x-axis at (a.sqrt3, 0) or (-a.sqrt3, 0).

   ✦ Working through each part gives: (0,a), (0,-a), and (a.sqrt3, 0) — or (-a.sqrt3, 0) for the mirror-image triangle.

3. 9.1.33 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the distance between P(x1,y1) and Q(x2,y2) when (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

   Hint. When a segment is parallel to an axis, one coordinate is identical for both points, so the distance formula collapses to a single absolute difference.

   (i) Parallel to the y-axis means x1=x2, so PQ = sqrt(0+(y2-y1)^2) = |y2-y1|. (ii) Parallel to the x-axis means y1=y2, so PQ = sqrt((x2-x1)^2+0) = |x2-x1|.

   ✦ Working through each part gives: (i) PQ = |y2-y1|. (ii) PQ = |x2-x1|.

4. 9.1.43 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find a point on the x-axis, which is equidistant from the points (7,6) and (3,4).

   Hint. Let the point be (x,0), set the two squared distances equal, and solve for x.

   Let the point be (x,0). (x-7)^2+36 = (x-3)^2+16. Expanding: x^2-14x+49+36 = x^2-6x+9+16, so -14x+85=-6x+25, giving -8x=-60, x=15/2.

   ✦ Working through each part gives: (15/2, 0).

5. 9.1.53 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the slope of a line which passes through the origin and the midpoint of the line segment joining P(0,-4) and B(8,0).

   Hint. Find the midpoint first, then compute the slope from the origin to that midpoint.

   Midpoint of P(0,-4) and B(8,0) is (4,-2). Slope from (0,0) to (4,-2) is (-2-0)/(4-0) = -1/2.

   ✦ Working through each part gives: -1/2.

6. 9.1.64 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Without using the Pythagoras theorem, show that the points (4,4), (3,5), and (-1,-1) are the vertices of a right angled triangle.

   Hint. Compute the three slopes and look for a pair whose product is -1, which identifies the right angle directly.

   Slope AB (A=(4,4),B=(3,5)) = (5-4)/(3-4) = -1. Slope AC (A=(4,4),C=(-1,-1)) = (-1-4)/(-1-4) = 1. Since slope(AB) x slope(AC) = -1 x 1 = -1, AB is perpendicular to AC, so the angle at A is 90 degrees.

   ✦ Working through each part gives: the triangle has a right angle at A, since slope(AB).slope(AC) = -1.

7. 9.1.73 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the slope of the line which makes an angle of 30 degrees with the positive direction of the y-axis, measured anticlockwise.

   Hint. An angle measured from the positive y-axis converts to an inclination (measured from the positive x-axis) by adding 90 degrees.

   An angle of 30 degrees from the positive y-axis (anticlockwise) corresponds to an inclination of 90+30=120 degrees from the positive x-axis. Slope = tan(120 degrees) = -tan(60 degrees) = -sqrt3.

   ✦ Working through each part gives: -sqrt3.

8. 9.1.84 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Without using the distance formula, show that the points (-2,-1), (4,0), (3,3), and (-3,2) are the vertices of a parallelogram.

   Hint. Use the property that the diagonals of a parallelogram bisect each other -- compute the midpoints of both diagonals and check they coincide.

   Taking the vertices in order A(-2,-1), B(4,0), C(3,3), D(-3,2), the diagonals are AC and BD. Midpoint of AC = ((-2+3)/2,(-1+3)/2) = (1/2,1). Midpoint of BD = ((4-3)/2,(0+2)/2) = (1/2,1). Since both midpoints coincide, the diagonals bisect each other, so ABCD is a parallelogram.

   ✦ Working through each part gives: proved: both diagonals share the midpoint (1/2, 1), so ABCD is a parallelogram.

9. 9.1.93 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

   Find the angle between the x-axis and the line joining the points (3,-1) and (4,-2).

   Hint. Compute the slope, then relate it to the acute angle the line makes with the x-axis.

   Slope = (-2-(-1))/(4-3) = -1/1 = -1. Since |slope|=1, the acute angle between the line and the x-axis is 45 degrees (the line's inclination itself is 135 degrees, but the angle it makes with the x-axis, taken as acute, is 45 degrees).

   ✦ Working through each part gives: 45 degrees.

10. 9.1.104 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    The slope of a line is double the slope of another line. If the tangent of the angle between them is 1/3, find the slopes of the lines.

    Hint. Let the slopes be m and 2m, substitute into the angle formula, and solve the resulting equation for both possible signs.

    tan(theta) = |(2m-m)/(1+2m.m)| = |m/(1+2m^2)| = 1/3, so 3|m|=1+2m^2 (since 1+2m^2 is always positive). For m>0: 3m=1+2m^2, giving 2m^2-3m+1=0, i.e. (2m-1)(m-1)=0, so m=1/2 or m=1. For m<0: -3m=1+2m^2, giving 2m^2+3m+1=0, i.e. (2m+1)(m+1)=0, so m=-1/2 or m=-1.

    ✦ Working through each part gives: the slope pairs (m, 2m) are (1/2, 1), (1, 2), (-1/2, -1), or (-1, -2).

11. 9.1.113 marksNCERT Class 11 Mathematics, Straight Lines, Reprint 2026-27

    A line passes through (x1,y1) and (h,k). If the slope of the line is m, show that k - y1 = m(h - x1).

    Hint. Write the slope formula for the two given points and set it equal to m, then cross-multiply.

    Since (x1,y1) and (h,k) both lie on the line, the slope between them equals m: (k-y1)/(h-x1) = m. Cross-multiplying gives k-y1 = m(h-x1).

    ✦ Working through each part gives: proved directly from the definition of slope: k - y1 = m(h - x1).

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.