JEE Mathematics Coordinates and Straight Lines MCQs Set 09

Answer: (a) The origin is the midpoint of the segment

Explanation:
1. The midpoint is \(\left(\frac{2-2}{2},\frac{4-4}{2}\right)=(0,0)\).
2. So the origin lies on the segment and is exactly its midpoint, dividing it in the ratio \(1:1\).

Teacher's Note:
1. The midpoint of \((x_1,y_1)\) and \((x_2,y_2)\) is \(\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\).
2. Points that are negatives of each other always have the origin as their midpoint.

Answer: (b) \(\frac{11}{8}\) sq. units

Explanation:
1. Area \(=\frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|\).
2. \(=\frac12\left|-2\left(0-\frac12\right)+5\left(\frac12-\frac34\right)+4\left(\frac34-0\right)\right|=\frac12\left|1-\frac54+3\right|\).
3. \(=\frac12\cdot\frac{11}{4}=\frac{11}{8}\) sq. units.

Teacher's Note:
1. Keep fractions as they are and add carefully; do not convert to decimals too early.
2. The absolute value is needed because the sign depends on the order of the vertices.

Answer: (a) the sum of the diagonals of \(ABCD\), \(2\sqrt{29}+2\sqrt{41}\)

Explanation:
1. Each side of the mid-point quadrilateral is parallel to a diagonal and equals half of it (mid-point theorem).
2. The four sides are \(\frac{AC}2,\frac{BD}2,\frac{AC}2,\frac{BD}2\), so the perimeter is \(AC+BD\).
3. \(AC=\sqrt{10^2+4^2}=2\sqrt{29}\) and \(BD=\sqrt{10^2+8^2}=2\sqrt{41}\), so the perimeter is \(2\sqrt{29}+2\sqrt{41}\).

Teacher's Note:
1. The mid-point quadrilateral of any quadrilateral is a parallelogram.
2. Its perimeter always equals the sum of the diagonals of the original quadrilateral.

Answer: (a) \(29\) sq. units

Explanation:
1. Area \(=\frac12|5(-3+5)+(-9)(-5-2)+(-3)(2+3)|\).
2. \(=\frac12|10+63-15|=\frac12\cdot58=29\) sq. units.

Teacher's Note:
1. Write the three terms separately before adding to avoid sign slips.
2. Forgetting the factor \(\frac12\) gives the area of the parallelogram, not the triangle.

Answer: (c) \(6x+5y=30\)

Explanation:
1. Intercept form: \(\frac x5+\frac y6=1\).
2. Multiplying by 30 gives \(6x+5y=30\).
3. Check: \((5,0)\) and \((0,6)\) both satisfy it.

Teacher's Note:
1. Intercept form \(\frac xa+\frac yb=1\) cuts off \(a\) on the x-axis and \(b\) on the y-axis.
2. The coefficient of \(x\) comes from the y-intercept and vice versa after clearing denominators.

Answer: (b) \(\left(-\frac53,2\right)\)

Explanation:
1. The point nearer to \((-3,4)\) divides the join from \((1,-2)\) in the ratio \(2:1\).
2. \(x=\frac{2(-3)+1(1)}{3}=-\frac53\).
3. \(y=\frac{2(4)+1(-2)}{3}=2\).

Teacher's Note:
1. The two trisection points divide the segment in the ratios \(1:2\) and \(2:1\).
2. Decide which end the point is closer to before choosing the ratio.

Answer: (a) \(x+y=7\)

Explanation:
1. \((x-2)^2+(y-3)^2=(x-4)^2+(y-5)^2\).
2. \(-4x-6y+13=-8x-10y+41\), so \(4x+4y=28\).
3. \(x+y=7\), which is the perpendicular bisector (midpoint \((3,4)\), slope \(-1\)).

Teacher's Note:
1. The locus is the perpendicular bisector of the join of the two points.
2. Check that the midpoint satisfies your equation.

Answer: (c) right-angled

Explanation:
1. The squared sides are \((-2,8)\)-\((-6,1)\): \(16+49=65\); \((-6,1)\)-\((0,4)\): \(36+9=45\); \((0,4)\)-\((-2,8)\): \(4+16=20\).
2. Since \(45+20=65\), the triangle is right-angled (the right angle is at \((0,4)\)).
3. No two sides are equal, so it is not isosceles.

Teacher's Note:
1. Test for a right angle with Pythagoras on the squared lengths.
2. Always compare the squares to avoid square roots.

Answer: (d) not a triangle, since the points are collinear

Explanation:
1. Area \(=\frac12\left|a[(c+a)-(a+b)]+b[(a+b)-(b+c)]+c[(b+c)-(c+a)]\right|\).
2. \(=\frac12|a(c-b)+b(a-c)+c(b-a)|=\frac12|ac-ab+ab-bc+bc-ac|=0\).
3. The three points are collinear (they all lie on the line \(x+y=a+b+c\)).

Teacher's Note:
1. Each point has coordinates adding up to \(a+b+c\).
2. That immediately shows they lie on the line \(x+y=a+b+c\).

Answer: (b) exactly two possible values, \(4\) and \(6\)

Explanation:
1. Area \(=\frac12|-2(p-9)+1(9-1)+4(1-p)|=\frac12|30-6p|\).
2. Setting this equal to \(3\) gives \(|30-6p|=6\).
3. So \(30-6p=\pm6\), giving \(p=4\) or \(p=6\).

Teacher's Note:
1. An area equation always has the modulus, so expect two cases.
2. Both values are valid because the area is positive in each.

Answer: (b) \(-1\)

Explanation:
1. Slope of the line through \((5,7)\) and \((8,11)\) is \(\frac{4}{3}\).
2. The slope from \((x,-1)\) to \((5,7)\) is \(\frac{8}{5-x}\).
3. Equating, \(\frac{8}{5-x}=\frac43\), so \(5-x=6\) and \(x=-1\).

Teacher's Note:
1. Collinear points have equal slopes between consecutive pairs.
2. The area method gives the same result.

Answer: (c) \(\sqrt{(a-b)^2+(b-c)^2}\)

Explanation:
1. The difference of x-coordinates is \((b+c)-(c+a)=b-a\).
2. The difference of y-coordinates is \((c+a)-(a+b)=c-b\).
3. So \(d=\sqrt{(b-a)^2+(c-b)^2}=\sqrt{(a-b)^2+(b-c)^2}\).

Teacher's Note:
1. The distance is always a sum of squares under the root, never a difference.
2. Squares make \((b-a)^2=(a-b)^2\).

Answer: (a) \((3,0)\)

Explanation:
1. Let the point be \((a,0)\). Then \((a-7)^2+36=(a+3)^2+16\).
2. \(a^2-14a+85=a^2+6a+25\), so \(20a=60\) and \(a=3\).
3. The point is \((3,0)\).

Teacher's Note:
1. A point on the x-axis has the form \((a,0)\) with only one unknown.
2. The \(a^2\) terms cancel, leaving a linear equation.