Exercise 9.1 answers: Straight Lines

Open NCERT p. 158Matches NCERT’s answer

1. Join the four points in order: A(−4, 5), B(0, 7), C(5, −5), D(−4, −2). This is quadrilateral ABCD.
2. Split it into two triangles by drawing the diagonal AC.
3. The area of a triangle with corners (x₁,y₁), (x₂,y₂), (x₃,y₃) is half of |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.
4. Area of triangle ABC = ½|−4(7−(−5)) + 0((−5)−5) + 5(5−7)| = ½|−48 + 0 −10| = 29 sq units.
5. Area of triangle ACD = ½|−4((−5)−(−2)) + 5((−2)−5) + (−4)(5−(−5))| = ½|12 −35 −40| = 63/2 sq units.
6. Add the two triangle areas: 29 + 63/2 = 121/2 sq units.

AnswerArea of the quadrilateral = 121/2 sq units (60.5 sq units).

Watch this explained “An area that is really a test”, 0:40 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 158Checked by computerAnswers can differ: one example

1. The base is on the y-axis with its mid-point at the origin, so the two base corners are (0, a) and (0, −a) — a distance 2a apart, matching the given side.
2. The third corner lies on the line through the base's mid-point, perpendicular to the base — here that line is the x-axis.
3. The height of an equilateral triangle of side 2a is (√3/2) × 2a = √3 a.
4. So the third vertex is at (√3 a, 0) — taking it on the other side of the y-axis, (−√3 a, 0), works equally well.

AnswerVertices: (0, a), (0, −a) and (√3 a, 0).

Open NCERT p. 159Matches NCERT’s answer

(i) PQ is parallel to the y-axis

1. When PQ is parallel to the y-axis, the x-coordinates are equal: x₁ = x₂
2. The distance formula gives: d = √((x₂ − x₁)² + (y₂ − y₁)²)
3. Since x₁ = x₂: d = √(0 + (y₂ − y₁)²) = |y₂ − y₁|

AnswerDistance = |y₂ − y₁|

(ii) PQ is parallel to the x-axis

1. When PQ is parallel to the x-axis, the y-coordinates are equal: y₁ = y₂
2. The distance formula gives: d = √((x₂ − x₁)² + (y₂ − y₁)²)
3. Since y₁ = y₂: d = √((x₂ − x₁)² + 0) = |x₂ − x₁|

AnswerDistance = |x₂ − x₁|

Watch the lesson Lines parallel to an axis, where one coordinate never changes

Open NCERT p. 159Matches NCERT’s answer

1. Let the point on the x-axis be (x, 0), since every point on the x-axis has y-coordinate 0.
2. Being equidistant from (7, 6) and (3, 4) means the squares of the two distances are equal: (x−7)² + 6² = (x−3)² + 4².
3. Expand both sides: x² −14x + 49 + 36 = x² −6x + 9 + 16.
4. Simplify: −14x + 85 = −6x + 25, so −8x = −60, giving x = 15/2.

AnswerThe point is (15/2, 0).

Open NCERT p. 159Matches NCERT’s answer

1. Find the mid-point of PB by averaging the x-coordinates and the y-coordinates: ((0+8)/2, (−4+0)/2) = (4, −2).
2. The line joins the origin (0, 0) to this mid-point (4, −2).
3. Slope = (change in y)/(change in x) = (−2 − 0)/(4 − 0) = −1/2.

AnswerSlope = −1/2.

Watch this explained “Where rise over run comes from”, 5:38 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 159One way to think about it

1. Call the points A(4, 4), B(3, 5), C(−1, −1).
2. Slope of AB = (5−4)/(3−4) = 1/(−1) = −1.
3. Slope of AC = (−1−4)/(−1−4) = (−5)/(−5) = 1.
4. Multiply the two slopes: (−1) × (1) = −1, so AB and AC are perpendicular.
5. A right angle at A, with AB ⊥ AC, makes ABC a right angled triangle — no distance formula needed.

In shortABC is right angled at A, since the slopes of AB and AC multiply to −1.

Watch this explained “Two shapes, and not one length”, 11:53 into Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular

Open NCERT p. 159Matches NCERT’s answer

1. Slope needs the angle measured from the positive x-axis, anticlockwise — that angle is called the line's inclination.
2. The positive y-axis is already 90° anticlockwise from the positive x-axis.
3. Sweeping a further 30° anticlockwise from there makes the inclination 90° + 30° = 120°.
4. Slope = tan(120°) = −tan(60°) = −√3.

AnswerSlope = −√3.

Watch this explained “Given an angle instead of a point”, 10:47 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 159One way to think about it

1. Call the points, in order, P(−2, −1), Q(4, 0), R(3, 3), S(−3, 2).
2. Slope of PQ = (0−(−1))/(4−(−2)) = 1/6.
3. Slope of RS = (2−3)/(−3−3) = (−1)/(−6) = 1/6, so PQ is parallel to RS.
4. Slope of QR = (3−0)/(3−4) = −3.
5. Slope of SP = (−1−2)/(−2−(−3)) = −3/1 = −3, so QR is parallel to SP.
6. Both pairs of opposite sides are parallel, so PQRS is a parallelogram.

In shortPQRS is a parallelogram, since PQ ∥ RS and QR ∥ SP (equal slopes).

Watch this explained “Two shapes, and not one length”, 11:53 into Equal slopes mean parallel; slopes multiplying to minus one mean perpendicular

Open NCERT p. 159Matches NCERT’s answer

1. Slope of the line = (−2−(−1))/(4−3) = −1/1 = −1.
2. The slope is negative, so the line falls to the right, and its inclination is an obtuse angle.
3. tan θ = −1 with 90° < θ < 180° gives θ = 180° − 45° = 135°.

AnswerThe angle is 135°.

Watch this explained “Given an angle instead of a point”, 10:47 into Steepness as the tangent of an angle, and the one line that has none

Open NCERT p. 159Checked by computerAnswers can differ: one example

1. Let the slopes be m and 2m.
2. The tangent of the angle θ between lines of slopes m₁ and m₂ is tan θ = |(m₂ − m₁)/(1 + m₁m₂)|.
3. So |(2m − m)/(1 + m × 2m)| = 1/3, that is |m|/(1 + 2m²) = 1/3, because 1 + 2m² is always positive.
4. If m is positive: 3m = 1 + 2m², so 2m² − 3m + 1 = 0, (2m − 1)(m − 1) = 0, and m = 1 or m = 1/2.
5. If m is negative: −3m = 1 + 2m², so 2m² + 3m + 1 = 0, (2m + 1)(m + 1) = 0, and m = −1 or m = −1/2.
6. Each value of m gives a pair m and 2m: 1 and 2, or 1/2 and 1, or −1 and −2, or −1/2 and −1.
7. Check one pair: slopes 1 and 2 give |(2 − 1)/(1 + 2)| = 1/3. ✓

AnswerThe slopes are 1 and 2. The question has four valid answers: 1 and 2, or 1/2 and 1, or −1 and −2, or −1/2 and −1.

Watch this explained “Four answers, not one”, 11:23 into Recovering the angle between two lines from their two slopes

Open NCERT p. 159One way to think about it

1. The line passes through both (x₁, y₁) and (h, k).
2. By the slope formula between two points on the same line, m = (k − y₁)/(h − x₁).
3. Multiply both sides by (h − x₁) to clear the fraction: m(h − x₁) = k − y₁.
4. So k − y₁ = m(h − x₁), as required.

In shortShown: k − y1 = m(h − x1), directly from the two-point slope formula.

Watch this explained “Multiplying through”, 2:15 into One point and a slope, or two points: the same condition written twice