MCQ
If the points $A(3,{\rm{ }}4),\,B(7,{\rm{ }}7),\,C(a,{\rm{ }}b)$ be collinear and $AC = 10$, then $(a,{\rm{ }}b)$=
- ✓$(11,{\rm{ }}10)$
- B$(10,{\rm{ }}11)$
- C$(11/2,\,5)$
- D$(5,{\rm{ }}11/2)$
✓
Answer
Correct option: A.
$(11,{\rm{ }}10)$
a
(a) ${(a - 3)^2} + {(b - 4)^2} = 100$ and $\frac{{b - 7}}{3} = \frac{{a - 7}}{4}$
(a) ${(a - 3)^2} + {(b - 4)^2} = 100$ and $\frac{{b - 7}}{3} = \frac{{a - 7}}{4}$
Hence $(a, b)$=$(11, 10)$.
Trick : Check with options. We find that the point $(11, 10)$ satisfies both the conditions i.e. $AC = \sqrt {{{(11 - 3)}^2} + {{(10 - 4)}^2}} = 10$.
Also this is collinear with $A,\,B$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If the equation $p{x^2} + (2 - q)xy + 3{y^2}$ $ - 6qx + 30y + 6q = 0$ represents a circle, then the values of $p$ and $q$ are
→The sum of integers from $1$ to $100$ that are divisible by $2$ or $5$ is
→Three letters are to be sent to different persons and addresses on the three envelopes are also written. Without looking at the addresses, the probability that the letters go into the right envelope is equal to
The total number of 9 - digit numbers which have all different digits is:
Let $F_1$ be the set of all parallelograms, $F_2$ the set of all rectangles, $F_3$ the set of all rhombuses, $F_4$ the set of all squares and $F_5$ the set of trapeziums in a plane. Then $F_1$ may be equal to:
→$ \text{f} (\text{x}) = \frac{\sqrt{(\text{x} 1) (\text{x} 3)}} {\text{(x} 2)}=$ is a real valued function in the domain:
→If $x$ be real, then the minimum value of ${x^2} - 8x + 17$ is
→The set $S = \left\{ {1,2,3, \ldots ,12} \right\}$ is to be partitioned into three sets $A,\,B,\, C$ of equal size . Thus $A \cup B \cup C = S$ અને $A \cap B = B \cap C = C \cap A = \emptyset $ . The number of ways to partition $S$ is
→An integer is chosen at random from the integers $\{1,2,3, \ldots \ldots . .50\}$. The probability that the chosen integer is a multiple of atleast one of $4,6$ and $7$ is
→Choose the correct answer. The inequality representing the following graph is:
